>>> x = {"a": 1, "b": 2}
>>> y = {"b": 10, "c": 11}
>>> z = x.update(y)
>>> print(z)
None
>>> x
{"a": 1, "b": 10, "c": 11}

d = {"x": 1, "y": 2, "z": 3} 
for key in d:
    print (key, "corresponds to", d[key])

Note: I have read Stack Overflow question here How do I sort a list of dictionaries by a value of the dictionary? and probably could change my code to have a list of dictionaries, but since I do not really need a list of dictionaries I wanted to know if there is a simpler solution to sort either in ascending or descending order.

How can I add new keys to a dictionary?

Is it possible to add a key to a Python dictionary after it has been created?

It doesn"t seem to have an .add() method.

Check if a given key already exists in a dictionary

I wanted to test if a key exists in a dictionary before updating the value for the key. I wrote the following code:

if "key1" in dict.keys():
  print "blah"
else:
  print "boo"

I think this is not the best way to accomplish this task. Is there a better way to test for a key in the dictionary?

How do I sort a list of dictionaries by a value of the dictionary?

I have a list of dictionaries and want each item to be sorted by a specific value.

Take into consideration the list:

[{"name":"Homer", "age":39}, {"name":"Bart", "age":10}]

When sorted by name, it should become:

[{"name":"Bart", "age":10}, {"name":"Homer", "age":39}]

How can I remove a key from a Python dictionary?

When deleting a key from a dictionary, I use:

if "key" in my_dict:
    del my_dict["key"]

Is there a one line way of doing this?

Delete an element from a dictionary

Is there a way to delete an item from a dictionary in Python?

Additionally, how can I delete an item from a dictionary to return a copy (i.e., not modifying the original)?

How do I convert two lists into a dictionary?

Question by Guido García

Imagine that you have the following list.

keys = ["name", "age", "food"]
values = ["Monty", 42, "spam"]

What is the simplest way to produce the following dictionary?

a_dict = {"name": "Monty", "age": 42, "food": "spam"}

Create a dictionary with list comprehension

I like the Python list comprehension syntax.

Can it be used to create dictionaries too? For example, by iterating over pairs of keys and values:

mydict = {(k,v) for (k,v) in blah blah blah}  # doesn"t work

Answer #1

In Python, what is the purpose of __slots__ and what are the cases one should avoid this?

TLDR:

The special attribute __slots__ allows you to explicitly state which instance attributes you expect your object instances to have, with the expected results:

1. faster attribute access.
2. space savings in memory.

The space savings is from

1. Storing value references in slots instead of __dict__.
2. Denying __dict__ and __weakref__ creation if parent classes deny them and you declare __slots__.

Quick Caveats

Small caveat, you should only declare a particular slot one time in an inheritance tree. For example:

class Base:
    __slots__ = "foo", "bar"

class Right(Base):
    __slots__ = "baz", 

class Wrong(Base):
    __slots__ = "foo", "bar", "baz"        # redundant foo and bar

Python doesn"t object when you get this wrong (it probably should), problems might not otherwise manifest, but your objects will take up more space than they otherwise should. Python 3.8:

>>> from sys import getsizeof
>>> getsizeof(Right()), getsizeof(Wrong())
(56, 72)

This is because the Base"s slot descriptor has a slot separate from the Wrong"s. This shouldn"t usually come up, but it could:

>>> w = Wrong()
>>> w.foo = "foo"
>>> Base.foo.__get__(w)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: foo
>>> Wrong.foo.__get__(w)
"foo"

The biggest caveat is for multiple inheritance - multiple "parent classes with nonempty slots" cannot be combined.

To accommodate this restriction, follow best practices: Factor out all but one or all parents" abstraction which their concrete class respectively and your new concrete class collectively will inherit from - giving the abstraction(s) empty slots (just like abstract base classes in the standard library).

See section on multiple inheritance below for an example.

Requirements:

• To have attributes named in __slots__ to actually be stored in slots instead of a __dict__, a class must inherit from object (automatic in Python 3, but must be explicit in Python 2).

• To prevent the creation of a __dict__, you must inherit from object and all classes in the inheritance must declare __slots__ and none of them can have a "__dict__" entry.

There are a lot of details if you wish to keep reading.

Why use __slots__: Faster attribute access.

The creator of Python, Guido van Rossum, states that he actually created __slots__ for faster attribute access.

It is trivial to demonstrate measurably significant faster access:

import timeit

class Foo(object): __slots__ = "foo",

class Bar(object): pass

slotted = Foo()
not_slotted = Bar()

def get_set_delete_fn(obj):
    def get_set_delete():
        obj.foo = "foo"
        obj.foo
        del obj.foo
    return get_set_delete

and

>>> min(timeit.repeat(get_set_delete_fn(slotted)))
0.2846834529991611
>>> min(timeit.repeat(get_set_delete_fn(not_slotted)))
0.3664822799983085

The slotted access is almost 30% faster in Python 3.5 on Ubuntu.

>>> 0.3664822799983085 / 0.2846834529991611
1.2873325658284342

In Python 2 on Windows I have measured it about 15% faster.

Why use __slots__: Memory Savings

Another purpose of __slots__ is to reduce the space in memory that each object instance takes up.

My own contribution to the documentation clearly states the reasons behind this:

The space saved over using __dict__ can be significant.

SQLAlchemy attributes a lot of memory savings to __slots__.

To verify this, using the Anaconda distribution of Python 2.7 on Ubuntu Linux, with guppy.hpy (aka heapy) and sys.getsizeof, the size of a class instance without __slots__ declared, and nothing else, is 64 bytes. That does not include the __dict__. Thank you Python for lazy evaluation again, the __dict__ is apparently not called into existence until it is referenced, but classes without data are usually useless. When called into existence, the __dict__ attribute is a minimum of 280 bytes additionally.

In contrast, a class instance with __slots__ declared to be () (no data) is only 16 bytes, and 56 total bytes with one item in slots, 64 with two.

For 64 bit Python, I illustrate the memory consumption in bytes in Python 2.7 and 3.6, for __slots__ and __dict__ (no slots defined) for each point where the dict grows in 3.6 (except for 0, 1, and 2 attributes):

       Python 2.7             Python 3.6
attrs  __slots__  __dict__*   __slots__  __dict__* | *(no slots defined)
none   16         56 + 272†   16         56 + 112† | †if __dict__ referenced
one    48         56 + 272    48         56 + 112
two    56         56 + 272    56         56 + 112
six    88         56 + 1040   88         56 + 152
11     128        56 + 1040   128        56 + 240
22     216        56 + 3344   216        56 + 408     
43     384        56 + 3344   384        56 + 752

So, in spite of smaller dicts in Python 3, we see how nicely __slots__ scale for instances to save us memory, and that is a major reason you would want to use __slots__.

Just for completeness of my notes, note that there is a one-time cost per slot in the class"s namespace of 64 bytes in Python 2, and 72 bytes in Python 3, because slots use data descriptors like properties, called "members".

>>> Foo.foo
<member "foo" of "Foo" objects>
>>> type(Foo.foo)
<class "member_descriptor">
>>> getsizeof(Foo.foo)
72

Demonstration of __slots__:

To deny the creation of a __dict__, you must subclass object. Everything subclasses object in Python 3, but in Python 2 you had to be explicit:

class Base(object): 
    __slots__ = ()

now:

>>> b = Base()
>>> b.a = "a"
Traceback (most recent call last):
  File "<pyshell#38>", line 1, in <module>
    b.a = "a"
AttributeError: "Base" object has no attribute "a"

Or subclass another class that defines __slots__

class Child(Base):
    __slots__ = ("a",)

and now:

c = Child()
c.a = "a"

but:

>>> c.b = "b"
Traceback (most recent call last):
  File "<pyshell#42>", line 1, in <module>
    c.b = "b"
AttributeError: "Child" object has no attribute "b"

To allow __dict__ creation while subclassing slotted objects, just add "__dict__" to the __slots__ (note that slots are ordered, and you shouldn"t repeat slots that are already in parent classes):

class SlottedWithDict(Child): 
    __slots__ = ("__dict__", "b")

swd = SlottedWithDict()
swd.a = "a"
swd.b = "b"
swd.c = "c"

and

>>> swd.__dict__
{"c": "c"}

Or you don"t even need to declare __slots__ in your subclass, and you will still use slots from the parents, but not restrict the creation of a __dict__:

class NoSlots(Child): pass
ns = NoSlots()
ns.a = "a"
ns.b = "b"

And:

>>> ns.__dict__
{"b": "b"}

However, __slots__ may cause problems for multiple inheritance:

class BaseA(object): 
    __slots__ = ("a",)

class BaseB(object): 
    __slots__ = ("b",)

Because creating a child class from parents with both non-empty slots fails:

>>> class Child(BaseA, BaseB): __slots__ = ()
Traceback (most recent call last):
  File "<pyshell#68>", line 1, in <module>
    class Child(BaseA, BaseB): __slots__ = ()
TypeError: Error when calling the metaclass bases
    multiple bases have instance lay-out conflict

If you run into this problem, You could just remove __slots__ from the parents, or if you have control of the parents, give them empty slots, or refactor to abstractions:

from abc import ABC

class AbstractA(ABC):
    __slots__ = ()

class BaseA(AbstractA): 
    __slots__ = ("a",)

class AbstractB(ABC):
    __slots__ = ()

class BaseB(AbstractB): 
    __slots__ = ("b",)

class Child(AbstractA, AbstractB): 
    __slots__ = ("a", "b")

c = Child() # no problem!

Add "__dict__" to __slots__ to get dynamic assignment:

class Foo(object):
    __slots__ = "bar", "baz", "__dict__"

and now:

>>> foo = Foo()
>>> foo.boink = "boink"

So with "__dict__" in slots we lose some of the size benefits with the upside of having dynamic assignment and still having slots for the names we do expect.

When you inherit from an object that isn"t slotted, you get the same sort of semantics when you use __slots__ - names that are in __slots__ point to slotted values, while any other values are put in the instance"s __dict__.

Avoiding __slots__ because you want to be able to add attributes on the fly is actually not a good reason - just add "__dict__" to your __slots__ if this is required.

You can similarly add __weakref__ to __slots__ explicitly if you need that feature.

Set to empty tuple when subclassing a namedtuple:

The namedtuple builtin make immutable instances that are very lightweight (essentially, the size of tuples) but to get the benefits, you need to do it yourself if you subclass them:

from collections import namedtuple
class MyNT(namedtuple("MyNT", "bar baz")):
    """MyNT is an immutable and lightweight object"""
    __slots__ = ()

usage:

>>> nt = MyNT("bar", "baz")
>>> nt.bar
"bar"
>>> nt.baz
"baz"

And trying to assign an unexpected attribute raises an AttributeError because we have prevented the creation of __dict__:

>>> nt.quux = "quux"
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: "MyNT" object has no attribute "quux"

You can allow __dict__ creation by leaving off __slots__ = (), but you can"t use non-empty __slots__ with subtypes of tuple.

Biggest Caveat: Multiple inheritance

Even when non-empty slots are the same for multiple parents, they cannot be used together:

class Foo(object): 
    __slots__ = "foo", "bar"
class Bar(object):
    __slots__ = "foo", "bar" # alas, would work if empty, i.e. ()

>>> class Baz(Foo, Bar): pass
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: Error when calling the metaclass bases
    multiple bases have instance lay-out conflict

Using an empty __slots__ in the parent seems to provide the most flexibility, allowing the child to choose to prevent or allow (by adding "__dict__" to get dynamic assignment, see section above) the creation of a __dict__:

class Foo(object): __slots__ = ()
class Bar(object): __slots__ = ()
class Baz(Foo, Bar): __slots__ = ("foo", "bar")
b = Baz()
b.foo, b.bar = "foo", "bar"

You don"t have to have slots - so if you add them, and remove them later, it shouldn"t cause any problems.

Going out on a limb here: If you"re composing mixins or using abstract base classes, which aren"t intended to be instantiated, an empty __slots__ in those parents seems to be the best way to go in terms of flexibility for subclassers.

To demonstrate, first, let"s create a class with code we"d like to use under multiple inheritance

class AbstractBase:
    __slots__ = ()
    def __init__(self, a, b):
        self.a = a
        self.b = b
    def __repr__(self):
        return f"{type(self).__name__}({repr(self.a)}, {repr(self.b)})"

We could use the above directly by inheriting and declaring the expected slots:

class Foo(AbstractBase):
    __slots__ = "a", "b"

But we don"t care about that, that"s trivial single inheritance, we need another class we might also inherit from, maybe with a noisy attribute:

class AbstractBaseC:
    __slots__ = ()
    @property
    def c(self):
        print("getting c!")
        return self._c
    @c.setter
    def c(self, arg):
        print("setting c!")
        self._c = arg

Now if both bases had nonempty slots, we couldn"t do the below. (In fact, if we wanted, we could have given AbstractBase nonempty slots a and b, and left them out of the below declaration - leaving them in would be wrong):

class Concretion(AbstractBase, AbstractBaseC):
    __slots__ = "a b _c".split()

And now we have functionality from both via multiple inheritance, and can still deny __dict__ and __weakref__ instantiation:

>>> c = Concretion("a", "b")
>>> c.c = c
setting c!
>>> c.c
getting c!
Concretion("a", "b")
>>> c.d = "d"
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: "Concretion" object has no attribute "d"

Other cases to avoid slots:

• Avoid them when you want to perform __class__ assignment with another class that doesn"t have them (and you can"t add them) unless the slot layouts are identical. (I am very interested in learning who is doing this and why.)
• Avoid them if you want to subclass variable length builtins like long, tuple, or str, and you want to add attributes to them.
• Avoid them if you insist on providing default values via class attributes for instance variables.

You may be able to tease out further caveats from the rest of the __slots__ documentation (the 3.7 dev docs are the most current), which I have made significant recent contributions to.

Critiques of other answers

The current top answers cite outdated information and are quite hand-wavy and miss the mark in some important ways.

Do not "only use __slots__ when instantiating lots of objects"

I quote:

"You would want to use __slots__ if you are going to instantiate a lot (hundreds, thousands) of objects of the same class."

Abstract Base Classes, for example, from the collections module, are not instantiated, yet __slots__ are declared for them.

Why?

If a user wishes to deny __dict__ or __weakref__ creation, those things must not be available in the parent classes.

__slots__ contributes to reusability when creating interfaces or mixins.

It is true that many Python users aren"t writing for reusability, but when you are, having the option to deny unnecessary space usage is valuable.

__slots__ doesn"t break pickling

When pickling a slotted object, you may find it complains with a misleading TypeError:

>>> pickle.loads(pickle.dumps(f))
TypeError: a class that defines __slots__ without defining __getstate__ cannot be pickled

This is actually incorrect. This message comes from the oldest protocol, which is the default. You can select the latest protocol with the -1 argument. In Python 2.7 this would be 2 (which was introduced in 2.3), and in 3.6 it is 4.

>>> pickle.loads(pickle.dumps(f, -1))
<__main__.Foo object at 0x1129C770>

in Python 2.7:

>>> pickle.loads(pickle.dumps(f, 2))
<__main__.Foo object at 0x1129C770>

in Python 3.6

>>> pickle.loads(pickle.dumps(f, 4))
<__main__.Foo object at 0x1129C770>

So I would keep this in mind, as it is a solved problem.

Critique of the (until Oct 2, 2016) accepted answer

The first paragraph is half short explanation, half predictive. Here"s the only part that actually answers the question

The proper use of __slots__ is to save space in objects. Instead of having a dynamic dict that allows adding attributes to objects at anytime, there is a static structure which does not allow additions after creation. This saves the overhead of one dict for every object that uses slots

The second half is wishful thinking, and off the mark:

While this is sometimes a useful optimization, it would be completely unnecessary if the Python interpreter was dynamic enough so that it would only require the dict when there actually were additions to the object.

Python actually does something similar to this, only creating the __dict__ when it is accessed, but creating lots of objects with no data is fairly ridiculous.

The second paragraph oversimplifies and misses actual reasons to avoid __slots__. The below is not a real reason to avoid slots (for actual reasons, see the rest of my answer above.):

They change the behavior of the objects that have slots in a way that can be abused by control freaks and static typing weenies.

It then goes on to discuss other ways of accomplishing that perverse goal with Python, not discussing anything to do with __slots__.

The third paragraph is more wishful thinking. Together it is mostly off-the-mark content that the answerer didn"t even author and contributes to ammunition for critics of the site.

Memory usage evidence

Create some normal objects and slotted objects:

>>> class Foo(object): pass
>>> class Bar(object): __slots__ = ()

Instantiate a million of them:

>>> foos = [Foo() for f in xrange(1000000)]
>>> bars = [Bar() for b in xrange(1000000)]

Inspect with guppy.hpy().heap():

>>> guppy.hpy().heap()
Partition of a set of 2028259 objects. Total size = 99763360 bytes.
 Index  Count   %     Size   % Cumulative  % Kind (class / dict of class)
     0 1000000  49 64000000  64  64000000  64 __main__.Foo
     1     169   0 16281480  16  80281480  80 list
     2 1000000  49 16000000  16  96281480  97 __main__.Bar
     3   12284   1   987472   1  97268952  97 str
...

Access the regular objects and their __dict__ and inspect again:

>>> for f in foos:
...     f.__dict__
>>> guppy.hpy().heap()
Partition of a set of 3028258 objects. Total size = 379763480 bytes.
 Index  Count   %      Size    % Cumulative  % Kind (class / dict of class)
     0 1000000  33 280000000  74 280000000  74 dict of __main__.Foo
     1 1000000  33  64000000  17 344000000  91 __main__.Foo
     2     169   0  16281480   4 360281480  95 list
     3 1000000  33  16000000   4 376281480  99 __main__.Bar
     4   12284   0    987472   0 377268952  99 str
...

This is consistent with the history of Python, from Unifying types and classes in Python 2.2

If you subclass a built-in type, extra space is automatically added to the instances to accomodate __dict__ and __weakrefs__. (The __dict__ is not initialized until you use it though, so you shouldn"t worry about the space occupied by an empty dictionary for each instance you create.) If you don"t need this extra space, you can add the phrase "__slots__ = []" to your class.

Answer #2

os.listdir() - list in the current directory

With listdir in os module you get the files and the folders in the current dir

 import os
 arr = os.listdir()
 print(arr)
 
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

Looking in a directory

arr = os.listdir("c:\files")

glob from glob

with glob you can specify a type of file to list like this

import glob

txtfiles = []
for file in glob.glob("*.txt"):
    txtfiles.append(file)

glob in a list comprehension

mylist = [f for f in glob.glob("*.txt")]

get the full path of only files in the current directory

import os
from os import listdir
from os.path import isfile, join

cwd = os.getcwd()
onlyfiles = [os.path.join(cwd, f) for f in os.listdir(cwd) if 
os.path.isfile(os.path.join(cwd, f))]
print(onlyfiles) 

["G:\getfilesname\getfilesname.py", "G:\getfilesname\example.txt"]

Getting the full path name with os.path.abspath

You get the full path in return

 import os
 files_path = [os.path.abspath(x) for x in os.listdir()]
 print(files_path)
 
 ["F:\documentiapplications.txt", "F:\documenticollections.txt"]

Walk: going through sub directories

os.walk returns the root, the directories list and the files list, that is why I unpacked them in r, d, f in the for loop; it, then, looks for other files and directories in the subfolders of the root and so on until there are no subfolders.

import os

# Getting the current work directory (cwd)
thisdir = os.getcwd()

# r=root, d=directories, f = files
for r, d, f in os.walk(thisdir):
    for file in f:
        if file.endswith(".docx"):
            print(os.path.join(r, file))

os.listdir(): get files in the current directory (Python 2)

In Python 2, if you want the list of the files in the current directory, you have to give the argument as "." or os.getcwd() in the os.listdir method.

 import os
 arr = os.listdir(".")
 print(arr)
 
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

To go up in the directory tree

# Method 1
x = os.listdir("..")

# Method 2
x= os.listdir("/")

Get files: os.listdir() in a particular directory (Python 2 and 3)

 import os
 arr = os.listdir("F:\python")
 print(arr)
 
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

Get files of a particular subdirectory with os.listdir()

import os

x = os.listdir("./content")

os.walk(".") - current directory

 import os
 arr = next(os.walk("."))[2]
 print(arr)
 
 >>> ["5bs_Turismo1.pdf", "5bs_Turismo1.pptx", "esperienza.txt"]

next(os.walk(".")) and os.path.join("dir", "file")

 import os
 arr = []
 for d,r,f in next(os.walk("F:\_python")):
     for file in f:
         arr.append(os.path.join(r,file))

 for f in arr:
     print(files)

>>> F:\_python\dict_class.py
>>> F:\_python\programmi.txt

next(os.walk("F:\") - get the full path - list comprehension

 [os.path.join(r,file) for r,d,f in next(os.walk("F:\_python")) for file in f]
 
 >>> ["F:\_python\dict_class.py", "F:\_python\programmi.txt"]

os.walk - get full path - all files in sub dirs**

x = [os.path.join(r,file) for r,d,f in os.walk("F:\_python") for file in f]
print(x)

>>> ["F:\_python\dict.py", "F:\_python\progr.txt", "F:\_python\readl.py"]

os.listdir() - get only txt files

 arr_txt = [x for x in os.listdir() if x.endswith(".txt")]
 print(arr_txt)
 
 >>> ["work.txt", "3ebooks.txt"]

Using glob to get the full path of the files

If I should need the absolute path of the files:

from path import path
from glob import glob
x = [path(f).abspath() for f in glob("F:\*.txt")]
for f in x:
    print(f)

>>> F:acquistionline.txt
>>> F:acquisti_2018.txt
>>> F:ootstrap_jquery_ecc.txt

Using os.path.isfile to avoid directories in the list

import os.path
listOfFiles = [f for f in os.listdir() if os.path.isfile(f)]
print(listOfFiles)

>>> ["a simple game.py", "data.txt", "decorator.py"]

Using pathlib from Python 3.4

import pathlib

flist = []
for p in pathlib.Path(".").iterdir():
    if p.is_file():
        print(p)
        flist.append(p)

 >>> error.PNG
 >>> exemaker.bat
 >>> guiprova.mp3
 >>> setup.py
 >>> speak_gui2.py
 >>> thumb.PNG

With list comprehension:

flist = [p for p in pathlib.Path(".").iterdir() if p.is_file()]

Alternatively, use pathlib.Path() instead of pathlib.Path(".")

Use glob method in pathlib.Path()

import pathlib

py = pathlib.Path().glob("*.py")
for file in py:
    print(file)

>>> stack_overflow_list.py
>>> stack_overflow_list_tkinter.py

Get all and only files with os.walk

import os
x = [i[2] for i in os.walk(".")]
y=[]
for t in x:
    for f in t:
        y.append(f)
print(y)

>>> ["append_to_list.py", "data.txt", "data1.txt", "data2.txt", "data_180617", "os_walk.py", "READ2.py", "read_data.py", "somma_defaltdic.py", "substitute_words.py", "sum_data.py", "data.txt", "data1.txt", "data_180617"]

Get only files with next and walk in a directory

 import os
 x = next(os.walk("F://python"))[2]
 print(x)
 
 >>> ["calculator.bat","calculator.py"]

Get only directories with next and walk in a directory

 import os
 next(os.walk("F://python"))[1] # for the current dir use (".")
 
 >>> ["python3","others"]

Get all the subdir names with walk

for r,d,f in os.walk("F:\_python"):
    for dirs in d:
        print(dirs)

>>> .vscode
>>> pyexcel
>>> pyschool.py
>>> subtitles
>>> _metaprogramming
>>> .ipynb_checkpoints

os.scandir() from Python 3.5 and greater

import os
x = [f.name for f in os.scandir() if f.is_file()]
print(x)

>>> ["calculator.bat","calculator.py"]

# Another example with scandir (a little variation from docs.python.org)
# This one is more efficient than os.listdir.
# In this case, it shows the files only in the current directory
# where the script is executed.

import os
with os.scandir() as i:
    for entry in i:
        if entry.is_file():
            print(entry.name)

>>> ebookmaker.py
>>> error.PNG
>>> exemaker.bat
>>> guiprova.mp3
>>> setup.py
>>> speakgui4.py
>>> speak_gui2.py
>>> speak_gui3.py
>>> thumb.PNG

Examples:

Ex. 1: How many files are there in the subdirectories?

In this example, we look for the number of files that are included in all the directory and its subdirectories.

import os

def count(dir, counter=0):
    "returns number of files in dir and subdirs"
    for pack in os.walk(dir):
        for f in pack[2]:
            counter += 1
    return dir + " : " + str(counter) + "files"

print(count("F:\python"))

>>> "F:\python" : 12057 files"

Ex.2: How to copy all files from a directory to another?

A script to make order in your computer finding all files of a type (default: pptx) and copying them in a new folder.

import os
import shutil
from path import path

destination = "F:\file_copied"
# os.makedirs(destination)

def copyfile(dir, filetype="pptx", counter=0):
    "Searches for pptx (or other - pptx is the default) files and copies them"
    for pack in os.walk(dir):
        for f in pack[2]:
            if f.endswith(filetype):
                fullpath = pack[0] + "\" + f
                print(fullpath)
                shutil.copy(fullpath, destination)
                counter += 1
    if counter > 0:
        print("-" * 30)
        print("	==> Found in: `" + dir + "` : " + str(counter) + " files
")

for dir in os.listdir():
    "searches for folders that starts with `_`"
    if dir[0] == "_":
        # copyfile(dir, filetype="pdf")
        copyfile(dir, filetype="txt")


>>> _compiti18Compito Contabilità 1conti.txt
>>> _compiti18Compito Contabilità 1modula4.txt
>>> _compiti18Compito Contabilità 1moduloa4.txt
>>> ------------------------
>>> ==> Found in: `_compiti18` : 3 files

Ex. 3: How to get all the files in a txt file

In case you want to create a txt file with all the file names:

import os
mylist = ""
with open("filelist.txt", "w", encoding="utf-8") as file:
    for eachfile in os.listdir():
        mylist += eachfile + "
"
    file.write(mylist)

Example: txt with all the files of an hard drive

"""
We are going to save a txt file with all the files in your directory.
We will use the function walk()
"""

import os

# see all the methods of os
# print(*dir(os), sep=", ")
listafile = []
percorso = []
with open("lista_file.txt", "w", encoding="utf-8") as testo:
    for root, dirs, files in os.walk("D:\"):
        for file in files:
            listafile.append(file)
            percorso.append(root + "\" + file)
            testo.write(file + "
")
listafile.sort()
print("N. of files", len(listafile))
with open("lista_file_ordinata.txt", "w", encoding="utf-8") as testo_ordinato:
    for file in listafile:
        testo_ordinato.write(file + "
")

with open("percorso.txt", "w", encoding="utf-8") as file_percorso:
    for file in percorso:
        file_percorso.write(file + "
")

os.system("lista_file.txt")
os.system("lista_file_ordinata.txt")
os.system("percorso.txt")

All the file of C: in one text file

This is a shorter version of the previous code. Change the folder where to start finding the files if you need to start from another position. This code generate a 50 mb on text file on my computer with something less then 500.000 lines with files with the complete path.

import os

with open("file.txt", "w", encoding="utf-8") as filewrite:
    for r, d, f in os.walk("C:\"):
        for file in f:
            filewrite.write(f"{r + file}
")

How to write a file with all paths in a folder of a type

With this function you can create a txt file that will have the name of a type of file that you look for (ex. pngfile.txt) with all the full path of all the files of that type. It can be useful sometimes, I think.

import os

def searchfiles(extension=".ttf", folder="H:\"):
    "Create a txt file with all the file of a type"
    with open(extension[1:] + "file.txt", "w", encoding="utf-8") as filewrite:
        for r, d, f in os.walk(folder):
            for file in f:
                if file.endswith(extension):
                    filewrite.write(f"{r + file}
")

# looking for png file (fonts) in the hard disk H:
searchfiles(".png", "H:\")

>>> H:4bs_18Dolphins5.png
>>> H:4bs_18Dolphins6.png
>>> H:4bs_18Dolphins7.png
>>> H:5_18marketing htmlassetsimageslogo2.png
>>> H:7z001.png
>>> H:7z002.png

(New) Find all files and open them with tkinter GUI

I just wanted to add in this 2019 a little app to search for all files in a dir and be able to open them by doubleclicking on the name of the file in the list.

import tkinter as tk
import os

def searchfiles(extension=".txt", folder="H:\"):
    "insert all files in the listbox"
    for r, d, f in os.walk(folder):
        for file in f:
            if file.endswith(extension):
                lb.insert(0, r + "\" + file)

def open_file():
    os.startfile(lb.get(lb.curselection()[0]))

root = tk.Tk()
root.geometry("400x400")
bt = tk.Button(root, text="Search", command=lambda:searchfiles(".png", "H:\"))
bt.pack()
lb = tk.Listbox(root)
lb.pack(fill="both", expand=1)
lb.bind("<Double-Button>", lambda x: open_file())
root.mainloop()

Answer #3

TL;DR: If you are using Python 3.10 or later, it just works. As of today (2019), in 3.7+ you must turn this feature on using a future statement (from __future__ import annotations). In Python 3.6 or below, use a string.

I guess you got this exception:

NameError: name "Position" is not defined

This is because Position must be defined before you can use it in an annotation unless you are using Python 3.10 or later.

Python 3.7+: from __future__ import annotations

Python 3.7 introduces PEP 563: postponed evaluation of annotations. A module that uses the future statement from __future__ import annotations will store annotations as strings automatically:

from __future__ import annotations

class Position:
    def __add__(self, other: Position) -> Position:
        ...

This is scheduled to become the default in Python 3.10. Since Python still is a dynamically typed language so no type checking is done at runtime, typing annotations should have no performance impact, right? Wrong! Before python 3.7 the typing module used to be one of the slowest python modules in core so if you import typing you will see up to 7 times increase in performance when you upgrade to 3.7.

Python <3.7: use a string

According to PEP 484, you should use a string instead of the class itself:

class Position:
    ...
    def __add__(self, other: "Position") -> "Position":
       ...

If you use the Django framework this may be familiar as Django models also use strings for forward references (foreign key definitions where the foreign model is self or is not declared yet). This should work with Pycharm and other tools.

Sources

The relevant parts of PEP 484 and PEP 563, to spare you the trip:

Forward references

When a type hint contains names that have not been defined yet, that definition may be expressed as a string literal, to be resolved later.

A situation where this occurs commonly is the definition of a container class, where the class being defined occurs in the signature of some of the methods. For example, the following code (the start of a simple binary tree implementation) does not work:

class Tree:
    def __init__(self, left: Tree, right: Tree):
        self.left = left
        self.right = right

To address this, we write:

class Tree:
    def __init__(self, left: "Tree", right: "Tree"):
        self.left = left
        self.right = right

The string literal should contain a valid Python expression (i.e., compile(lit, "", "eval") should be a valid code object) and it should evaluate without errors once the module has been fully loaded. The local and global namespace in which it is evaluated should be the same namespaces in which default arguments to the same function would be evaluated.

and PEP 563:

Implementation

In Python 3.10, function and variable annotations will no longer be evaluated at definition time. Instead, a string form will be preserved in the respective¬†__annotations__¬†dictionary. Static type checkers will see no difference in behavior, whereas tools using annotations at runtime will have to perform postponed evaluation.

...

Enabling the future behavior in Python 3.7

The functionality described above can be enabled starting from Python 3.7 using the following special import:

from __future__ import annotations

Things that you may be tempted to do instead

A. Define a dummy Position

Before the class definition, place a dummy definition:

class Position(object):
    pass


class Position(object):
    ...

This will get rid of the NameError and may even look OK:

>>> Position.__add__.__annotations__
{"other": __main__.Position, "return": __main__.Position}

But is it?

>>> for k, v in Position.__add__.__annotations__.items():
...     print(k, "is Position:", v is Position)                                                                                                                                                                                                                  
return is Position: False
other is Position: False

B. Monkey-patch in order to add the annotations:

You may want to try some Python meta programming magic and write a decorator to monkey-patch the class definition in order to add annotations:

class Position:
    ...
    def __add__(self, other):
        return self.__class__(self.x + other.x, self.y + other.y)

The decorator should be responsible for the equivalent of this:

Position.__add__.__annotations__["return"] = Position
Position.__add__.__annotations__["other"] = Position

At least it seems right:

>>> for k, v in Position.__add__.__annotations__.items():
...     print(k, "is Position:", v is Position)                                                                                                                                                                                                                  
return is Position: True
other is Position: True

Probably too much trouble.

Answer #4

As you are in python3 , use dict.items() instead of dict.iteritems()

iteritems() was removed in python3, so you can"t use this method anymore.

Take a look at Python 3.0 Wiki Built-in Changes section, where it is stated:

Removed dict.iteritems(), dict.iterkeys(), and dict.itervalues().

Instead: use dict.items(), dict.keys(), and dict.values() respectively.

Answer #5

My quick & dirty JSON dump that eats dates and everything:

json.dumps(my_dictionary, indent=4, sort_keys=True, default=str)

default is a function applied to objects that aren"t serializable.
In this case it"s str, so it just converts everything it doesn"t know to strings. Which is great for serialization but not so great when deserializing (hence the "quick & dirty") as anything might have been string-ified without warning, e.g. a function or numpy array.

Answer #6

a = numpy.array([0, 3, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 4])
unique, counts = numpy.unique(a, return_counts=True)
dict(zip(unique, counts))

# {0: 7, 1: 4, 2: 1, 3: 2, 4: 1}

Non-numpy way:

Use collections.Counter;

import collections, numpy
a = numpy.array([0, 3, 0, 1, 0, 1, 2, 1, 0, 0, 0, 0, 1, 3, 4])
collections.Counter(a)

# Counter({0: 7, 1: 4, 3: 2, 2: 1, 4: 1})

Answer #7

Because [] and {} are literal syntax. Python can create bytecode just to create the list or dictionary objects:

>>> import dis
>>> dis.dis(compile("[]", "", "eval"))
  1           0 BUILD_LIST               0
              3 RETURN_VALUE        
>>> dis.dis(compile("{}", "", "eval"))
  1           0 BUILD_MAP                0
              3 RETURN_VALUE        

list() and dict() are separate objects. Their names need to be resolved, the stack has to be involved to push the arguments, the frame has to be stored to retrieve later, and a call has to be made. That all takes more time.

For the empty case, that means you have at the very least a LOAD_NAME (which has to search through the global namespace as well as the builtins module) followed by a CALL_FUNCTION, which has to preserve the current frame:

>>> dis.dis(compile("list()", "", "eval"))
  1           0 LOAD_NAME                0 (list)
              3 CALL_FUNCTION            0
              6 RETURN_VALUE        
>>> dis.dis(compile("dict()", "", "eval"))
  1           0 LOAD_NAME                0 (dict)
              3 CALL_FUNCTION            0
              6 RETURN_VALUE        

You can time the name lookup separately with timeit:

>>> import timeit
>>> timeit.timeit("list", number=10**7)
0.30749011039733887
>>> timeit.timeit("dict", number=10**7)
0.4215109348297119

The time discrepancy there is probably a dictionary hash collision. Subtract those times from the times for calling those objects, and compare the result against the times for using literals:

>>> timeit.timeit("[]", number=10**7)
0.30478692054748535
>>> timeit.timeit("{}", number=10**7)
0.31482696533203125
>>> timeit.timeit("list()", number=10**7)
0.9991960525512695
>>> timeit.timeit("dict()", number=10**7)
1.0200958251953125

So having to call the object takes an additional 1.00 - 0.31 - 0.30 == 0.39 seconds per 10 million calls.

You can avoid the global lookup cost by aliasing the global names as locals (using a timeit setup, everything you bind to a name is a local):

>>> timeit.timeit("_list", "_list = list", number=10**7)
0.1866450309753418
>>> timeit.timeit("_dict", "_dict = dict", number=10**7)
0.19016098976135254
>>> timeit.timeit("_list()", "_list = list", number=10**7)
0.841480016708374
>>> timeit.timeit("_dict()", "_dict = dict", number=10**7)
0.7233691215515137

but you never can overcome that CALL_FUNCTION cost.

Answer #8

There are many ways to convert an instance to a dictionary, with varying degrees of corner case handling and closeness to the desired result.

1. instance.__dict__

instance.__dict__

which returns

{"_foreign_key_cache": <OtherModel: OtherModel object>,
 "_state": <django.db.models.base.ModelState at 0x7ff0993f6908>,
 "auto_now_add": datetime.datetime(2018, 12, 20, 21, 34, 29, 494827, tzinfo=<UTC>),
 "foreign_key_id": 2,
 "id": 1,
 "normal_value": 1,
 "readonly_value": 2}

This is by far the simplest, but is missing many_to_many, foreign_key is misnamed, and it has two unwanted extra things in it.

2. model_to_dict

from django.forms.models import model_to_dict
model_to_dict(instance)

which returns

{"foreign_key": 2,
 "id": 1,
 "many_to_many": [<OtherModel: OtherModel object>],
 "normal_value": 1}

This is the only one with many_to_many, but is missing the uneditable fields.

3. model_to_dict(..., fields=...)

from django.forms.models import model_to_dict
model_to_dict(instance, fields=[field.name for field in instance._meta.fields])

which returns

{"foreign_key": 2, "id": 1, "normal_value": 1}

This is strictly worse than the standard model_to_dict invocation.

4. query_set.values()

SomeModel.objects.filter(id=instance.id).values()[0]

which returns

{"auto_now_add": datetime.datetime(2018, 12, 20, 21, 34, 29, 494827, tzinfo=<UTC>),
 "foreign_key_id": 2,
 "id": 1,
 "normal_value": 1,
 "readonly_value": 2}

This is the same output as instance.__dict__ but without the extra fields. foreign_key_id is still wrong and many_to_many is still missing.

5. Custom Function

The code for django"s model_to_dict had most of the answer. It explicitly removed non-editable fields, so removing that check and getting the ids of foreign keys for many to many fields results in the following code which behaves as desired:

from itertools import chain

def to_dict(instance):
    opts = instance._meta
    data = {}
    for f in chain(opts.concrete_fields, opts.private_fields):
        data[f.name] = f.value_from_object(instance)
    for f in opts.many_to_many:
        data[f.name] = [i.id for i in f.value_from_object(instance)]
    return data

While this is the most complicated option, calling to_dict(instance) gives us exactly the desired result:

{"auto_now_add": datetime.datetime(2018, 12, 20, 21, 34, 29, 494827, tzinfo=<UTC>),
 "foreign_key": 2,
 "id": 1,
 "many_to_many": [2],
 "normal_value": 1,
 "readonly_value": 2}

6. Use Serializers

Django Rest Framework"s ModelSerialzer allows you to build a serializer automatically from a model.

from rest_framework import serializers
class SomeModelSerializer(serializers.ModelSerializer):
    class Meta:
        model = SomeModel
        fields = "__all__"

SomeModelSerializer(instance).data

returns

{"auto_now_add": "2018-12-20T21:34:29.494827Z",
 "foreign_key": 2,
 "id": 1,
 "many_to_many": [2],
 "normal_value": 1,
 "readonly_value": 2}

This is almost as good as the custom function, but auto_now_add is a string instead of a datetime object.

Bonus Round: better model printing

If you want a django model that has a better python command-line display, have your models child-class the following:

from django.db import models
from itertools import chain

class PrintableModel(models.Model):
    def __repr__(self):
        return str(self.to_dict())

    def to_dict(instance):
        opts = instance._meta
        data = {}
        for f in chain(opts.concrete_fields, opts.private_fields):
            data[f.name] = f.value_from_object(instance)
        for f in opts.many_to_many:
            data[f.name] = [i.id for i in f.value_from_object(instance)]
        return data

    class Meta:
        abstract = True

So, for example, if we define our models as such:

class OtherModel(PrintableModel): pass

class SomeModel(PrintableModel):
    normal_value = models.IntegerField()
    readonly_value = models.IntegerField(editable=False)
    auto_now_add = models.DateTimeField(auto_now_add=True)
    foreign_key = models.ForeignKey(OtherModel, related_name="ref1")
    many_to_many = models.ManyToManyField(OtherModel, related_name="ref2")

Calling SomeModel.objects.first() now gives output like this:

{"auto_now_add": datetime.datetime(2018, 12, 20, 21, 34, 29, 494827, tzinfo=<UTC>),
 "foreign_key": 2,
 "id": 1,
 "many_to_many": [2],
 "normal_value": 1,
 "readonly_value": 2}

Answer #9

Are dictionaries ordered in Python 3.6+?

They are insertion ordered^[1]. As of Python 3.6, for the CPython implementation of Python, dictionaries remember the order of items inserted. This is considered an implementation detail in Python 3.6; you need to use OrderedDict if you want insertion ordering that"s guaranteed across other implementations of Python (and other ordered behavior^[1]).

As of Python 3.7, this is no longer an implementation detail and instead becomes a language feature. From a python-dev message by GvR:

Make it so. "Dict keeps insertion order" is the ruling. Thanks!

This simply means that you can depend on it. Other implementations of Python must also offer an insertion ordered dictionary if they wish to be a conforming implementation of Python 3.7.

How does the Python 3.6 dictionary implementation perform better^[2] than the older one while preserving element order?

Essentially, by keeping two arrays.

• The first array, dk_entries, holds the entries (of type PyDictKeyEntry) for the dictionary in the order that they were inserted. Preserving order is achieved by this being an append only array where new items are always inserted at the end (insertion order).

• The second, dk_indices, holds the indices for the dk_entries array (that is, values that indicate the position of the corresponding entry in dk_entries). This array acts as the hash table. When a key is hashed it leads to one of the indices stored in dk_indices and the corresponding entry is fetched by indexing dk_entries. Since only indices are kept, the type of this array depends on the overall size of the dictionary (ranging from type int8_t(1 byte) to int32_t/int64_t (4/8 bytes) on 32/64 bit builds)

In the previous implementation, a sparse array of type PyDictKeyEntry and size dk_size had to be allocated; unfortunately, it also resulted in a lot of empty space since that array was not allowed to be more than 2/3 * dk_size full for performance reasons. (and the empty space still had PyDictKeyEntry size!).

This is not the case now since only the required entries are stored (those that have been inserted) and a sparse array of type intX_t (X depending on dict size) 2/3 * dk_sizes full is kept. The empty space changed from type PyDictKeyEntry to intX_t.

So, obviously, creating a sparse array of type PyDictKeyEntry is much more memory demanding than a sparse array for storing ints.

You can see the full conversation on Python-Dev regarding this feature if interested, it is a good read.

In the original proposal made by Raymond Hettinger, a visualization of the data structures used can be seen which captures the gist of the idea.

For example, the dictionary:

d = {"timmy": "red", "barry": "green", "guido": "blue"}

is currently stored as [keyhash, key, value]:

entries = [["--", "--", "--"],
           [-8522787127447073495, "barry", "green"],
           ["--", "--", "--"],
           ["--", "--", "--"],
           ["--", "--", "--"],
           [-9092791511155847987, "timmy", "red"],
           ["--", "--", "--"],
           [-6480567542315338377, "guido", "blue"]]

Instead, the data should be organized as follows:

indices =  [None, 1, None, None, None, 0, None, 2]
entries =  [[-9092791511155847987, "timmy", "red"],
            [-8522787127447073495, "barry", "green"],
            [-6480567542315338377, "guido", "blue"]]

As you can visually now see, in the original proposal, a lot of space is essentially empty to reduce collisions and make look-ups faster. With the new approach, you reduce the memory required by moving the sparseness where it"s really required, in the indices.

Answer #10

json.dumps() converts a dictionary to str object, not a json(dict) object! So you have to load your str into a dict to use it by using json.loads() method

See json.dumps() as a save method and json.loads() as a retrieve method.

This is the code sample which might help you understand it more:

import json

r = {"is_claimed": "True", "rating": 3.5}
r = json.dumps(r)
loaded_r = json.loads(r)
loaded_r["rating"] #Output 3.5
type(r) #Output str
type(loaded_r) #Output dict

Python program to find the sum of all elements in a dictionary: StackOverflow Questions

Finding the index of an item in a list

Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python?

Find current directory and file"s directory

In Python, what commands can I use to find:

1. the current directory (where I was in the terminal when I ran the Python script), and
2. where the file I am executing is?

How to find if directory exists in Python

In the os module in Python, is there a way to find if a directory exists, something like:

>>> os.direxists(os.path.join(os.getcwd()), "new_folder")) # in pseudocode
True/False

How do I find the location of my Python site-packages directory?

Question by Daryl Spitzer

How do I find the location of my site-packages directory?

Find all files in a directory with extension .txt in Python

How can I find all the files in a directory having the extension .txt in python?

Find which version of package is installed with pip

Using pip, is it possible to figure out which version of a package is currently installed?

I know about pip install XYZ --upgrade but I am wondering if there is anything like pip info XYZ. If not what would be the best way to tell what version I am currently using.

error: Unable to find vcvarsall.bat

I tried to install the Python package dulwich:

pip install dulwich

But I get a cryptic error message:

error: Unable to find vcvarsall.bat

The same happens if I try installing the package manually:

> python setup.py install
running build_ext
building "dulwich._objects" extension
error: Unable to find vcvarsall.bat

How to use glob() to find files recursively?

This is what I have:

glob(os.path.join("src","*.c"))

but I want to search the subfolders of src. Something like this would work:

glob(os.path.join("src","*.c"))
glob(os.path.join("src","*","*.c"))
glob(os.path.join("src","*","*","*.c"))
glob(os.path.join("src","*","*","*","*.c"))

But this is obviously limited and clunky.

Python: Find in list

I have come across this:

item = someSortOfSelection()
if item in myList:
    doMySpecialFunction(item)

but sometimes it does not work with all my items, as if they weren"t recognized in the list (when it"s a list of string).

Is this the most "pythonic" way of finding an item in a list: if x in l:?

How to find out the number of CPUs using python

I want to know the number of CPUs on the local machine using Python. The result should be user/real as output by time(1) when called with an optimally scaling userspace-only program.

Answer #1

How to iterate over rows in a DataFrame in Pandas?

Answer: DON"T^*!

Iteration in Pandas is an anti-pattern and is something you should only do when you have exhausted every other option. You should not use any function with "iter" in its name for more than a few thousand rows or you will have to get used to a lot of waiting.

Do you want to print a DataFrame? Use DataFrame.to_string().

Do you want to compute something? In that case, search for methods in this order (list modified from here):

1. Vectorization
2. Cython routines
3. List Comprehensions (vanilla for loop)
4. DataFrame.apply(): i) ¬†Reductions that can be performed in Cython, ii) Iteration in Python space
5. DataFrame.itertuples() and iteritems()
6. DataFrame.iterrows()

iterrows and itertuples (both receiving many votes in answers to this question) should be used in very rare circumstances, such as generating row objects/nametuples for sequential processing, which is really the only thing these functions are useful for.

Appeal to Authority

The documentation page on iteration has a huge red warning box that says:

Iterating through pandas objects is generally slow. In many cases, iterating manually over the rows is not needed [...].

_{* It"s actually a little more complicated than "don"t". df.iterrows() is the correct answer to this question, but "vectorize your ops" is the better one. I will concede that there are circumstances where iteration cannot be avoided (for example, some operations where the result depends on the value computed for the previous row). However, it takes some familiarity with the library to know when. If you"re not sure whether you need an iterative solution, you probably don"t. PS: To know more about my rationale for writing this answer, skip to the very bottom.}

Faster than Looping: Vectorization, Cython

A good number of basic operations and computations are "vectorised" by pandas (either through NumPy, or through Cythonized functions). This includes arithmetic, comparisons, (most) reductions, reshaping (such as pivoting), joins, and groupby operations. Look through the documentation on Essential Basic Functionality to find a suitable vectorised method for your problem.

If none exists, feel free to write your own using custom Cython extensions.

Next Best Thing: List Comprehensions^*

List comprehensions should be your next port of call if 1) there is no vectorized solution available, 2) performance is important, but not important enough to go through the hassle of cythonizing your code, and 3) you"re trying to perform elementwise transformation on your code. There is a good amount of evidence to suggest that list comprehensions are sufficiently fast (and even sometimes faster) for many common Pandas tasks.

The formula is simple,

# Iterating over one column - `f` is some function that processes your data
result = [f(x) for x in df["col"]]
# Iterating over two columns, use `zip`
result = [f(x, y) for x, y in zip(df["col1"], df["col2"])]
# Iterating over multiple columns - same data type
result = [f(row[0], ..., row[n]) for row in df[["col1", ...,"coln"]].to_numpy()]
# Iterating over multiple columns - differing data type
result = [f(row[0], ..., row[n]) for row in zip(df["col1"], ..., df["coln"])]

If you can encapsulate your business logic into a function, you can use a list comprehension that calls it. You can make arbitrarily complex things work through the simplicity and speed of raw Python code.

Caveats

List comprehensions assume that your data is easy to work with - what that means is your data types are consistent and you don"t have NaNs, but this cannot always be guaranteed.

1. The first one is more obvious, but when dealing with NaNs, prefer in-built pandas methods if they exist (because they have much better corner-case handling logic), or ensure your business logic includes appropriate NaN handling logic.
2. When dealing with mixed data types you should iterate over zip(df["A"], df["B"], ...) instead of df[["A", "B"]].to_numpy() as the latter implicitly upcasts data to the most common type. As an example if A is numeric and B is string, to_numpy() will cast the entire array to string, which may not be what you want. Fortunately zipping your columns together is the most straightforward workaround to this.

_{*Your mileage may vary for the reasons outlined in the Caveats section above.}

An Obvious Example

Let"s demonstrate the difference with a simple example of adding two pandas columns A + B. This is a vectorizable operaton, so it will be easy to contrast the performance of the methods discussed above.

Benchmarking code, for your reference. The line at the bottom measures a function written in numpandas, a style of Pandas that mixes heavily with NumPy to squeeze out maximum performance. Writing numpandas code should be avoided unless you know what you"re doing. Stick to the API where you can (i.e., prefer vec over vec_numpy).

I should mention, however, that it isn"t always this cut and dry. Sometimes the answer to "what is the best method for an operation" is "it depends on your data". My advice is to test out different approaches on your data before settling on one.

Further Reading

• 10 Minutes to pandas, and Essential Basic Functionality - Useful links that introduce you to Pandas and its library of vectorized*/cythonized functions.

• Enhancing Performance - A primer from the documentation on enhancing standard Pandas operations

• Are for-loops in pandas really bad? When should I care? - a detailed writeup by me on list comprehensions and their suitability for various operations (mainly ones involving non-numeric data)

• When should I (not) want to use pandas apply() in my code? - apply is slow (but not as slow as the iter* family. There are, however, situations where one can (or should) consider apply as a serious alternative, especially in some GroupBy operations).

_{* Pandas string methods are "vectorized" in the sense that they are specified on the series but operate on each element. The underlying mechanisms are still iterative, because string operations are inherently hard to vectorize.}

Why I Wrote this Answer

A common trend I notice from new users is to ask questions of the form "How can I iterate over my df to do X?". Showing code that calls iterrows() while doing something inside a for loop. Here is why. A new user to the library who has not been introduced to the concept of vectorization will likely envision the code that solves their problem as iterating over their data to do something. Not knowing how to iterate over a DataFrame, the first thing they do is Google it and end up here, at this question. They then see the accepted answer telling them how to, and they close their eyes and run this code without ever first questioning if iteration is not the right thing to do.

The aim of this answer is to help new users understand that iteration is not necessarily the solution to every problem, and that better, faster and more idiomatic solutions could exist, and that it is worth investing time in exploring them. I"m not trying to start a war of iteration vs. vectorization, but I want new users to be informed when developing solutions to their problems with this library.

Answer #2

In Python, what is the purpose of __slots__ and what are the cases one should avoid this?

TLDR:

The special attribute __slots__ allows you to explicitly state which instance attributes you expect your object instances to have, with the expected results:

1. faster attribute access.
2. space savings in memory.

The space savings is from

1. Storing value references in slots instead of __dict__.
2. Denying __dict__ and __weakref__ creation if parent classes deny them and you declare __slots__.

Quick Caveats

Small caveat, you should only declare a particular slot one time in an inheritance tree. For example:

class Base:
    __slots__ = "foo", "bar"

class Right(Base):
    __slots__ = "baz", 

class Wrong(Base):
    __slots__ = "foo", "bar", "baz"        # redundant foo and bar

Python doesn"t object when you get this wrong (it probably should), problems might not otherwise manifest, but your objects will take up more space than they otherwise should. Python 3.8:

>>> from sys import getsizeof
>>> getsizeof(Right()), getsizeof(Wrong())
(56, 72)

This is because the Base"s slot descriptor has a slot separate from the Wrong"s. This shouldn"t usually come up, but it could:

>>> w = Wrong()
>>> w.foo = "foo"
>>> Base.foo.__get__(w)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: foo
>>> Wrong.foo.__get__(w)
"foo"

The biggest caveat is for multiple inheritance - multiple "parent classes with nonempty slots" cannot be combined.

To accommodate this restriction, follow best practices: Factor out all but one or all parents" abstraction which their concrete class respectively and your new concrete class collectively will inherit from - giving the abstraction(s) empty slots (just like abstract base classes in the standard library).

See section on multiple inheritance below for an example.

Requirements:

• To have attributes named in __slots__ to actually be stored in slots instead of a __dict__, a class must inherit from object (automatic in Python 3, but must be explicit in Python 2).

• To prevent the creation of a __dict__, you must inherit from object and all classes in the inheritance must declare __slots__ and none of them can have a "__dict__" entry.

There are a lot of details if you wish to keep reading.

Why use __slots__: Faster attribute access.

The creator of Python, Guido van Rossum, states that he actually created __slots__ for faster attribute access.

It is trivial to demonstrate measurably significant faster access:

import timeit

class Foo(object): __slots__ = "foo",

class Bar(object): pass

slotted = Foo()
not_slotted = Bar()

def get_set_delete_fn(obj):
    def get_set_delete():
        obj.foo = "foo"
        obj.foo
        del obj.foo
    return get_set_delete

and

>>> min(timeit.repeat(get_set_delete_fn(slotted)))
0.2846834529991611
>>> min(timeit.repeat(get_set_delete_fn(not_slotted)))
0.3664822799983085

The slotted access is almost 30% faster in Python 3.5 on Ubuntu.

>>> 0.3664822799983085 / 0.2846834529991611
1.2873325658284342

In Python 2 on Windows I have measured it about 15% faster.

Why use __slots__: Memory Savings

Another purpose of __slots__ is to reduce the space in memory that each object instance takes up.

My own contribution to the documentation clearly states the reasons behind this:

The space saved over using __dict__ can be significant.

SQLAlchemy attributes a lot of memory savings to __slots__.

To verify this, using the Anaconda distribution of Python 2.7 on Ubuntu Linux, with guppy.hpy (aka heapy) and sys.getsizeof, the size of a class instance without __slots__ declared, and nothing else, is 64 bytes. That does not include the __dict__. Thank you Python for lazy evaluation again, the __dict__ is apparently not called into existence until it is referenced, but classes without data are usually useless. When called into existence, the __dict__ attribute is a minimum of 280 bytes additionally.

In contrast, a class instance with __slots__ declared to be () (no data) is only 16 bytes, and 56 total bytes with one item in slots, 64 with two.

For 64 bit Python, I illustrate the memory consumption in bytes in Python 2.7 and 3.6, for __slots__ and __dict__ (no slots defined) for each point where the dict grows in 3.6 (except for 0, 1, and 2 attributes):

       Python 2.7             Python 3.6
attrs  __slots__  __dict__*   __slots__  __dict__* | *(no slots defined)
none   16         56 + 272†   16         56 + 112† | †if __dict__ referenced
one    48         56 + 272    48         56 + 112
two    56         56 + 272    56         56 + 112
six    88         56 + 1040   88         56 + 152
11     128        56 + 1040   128        56 + 240
22     216        56 + 3344   216        56 + 408     
43     384        56 + 3344   384        56 + 752

So, in spite of smaller dicts in Python 3, we see how nicely __slots__ scale for instances to save us memory, and that is a major reason you would want to use __slots__.

Just for completeness of my notes, note that there is a one-time cost per slot in the class"s namespace of 64 bytes in Python 2, and 72 bytes in Python 3, because slots use data descriptors like properties, called "members".

>>> Foo.foo
<member "foo" of "Foo" objects>
>>> type(Foo.foo)
<class "member_descriptor">
>>> getsizeof(Foo.foo)
72

Demonstration of __slots__:

To deny the creation of a __dict__, you must subclass object. Everything subclasses object in Python 3, but in Python 2 you had to be explicit:

class Base(object): 
    __slots__ = ()

now:

>>> b = Base()
>>> b.a = "a"
Traceback (most recent call last):
  File "<pyshell#38>", line 1, in <module>
    b.a = "a"
AttributeError: "Base" object has no attribute "a"

Or subclass another class that defines __slots__

class Child(Base):
    __slots__ = ("a",)

and now:

c = Child()
c.a = "a"

but:

>>> c.b = "b"
Traceback (most recent call last):
  File "<pyshell#42>", line 1, in <module>
    c.b = "b"
AttributeError: "Child" object has no attribute "b"

To allow __dict__ creation while subclassing slotted objects, just add "__dict__" to the __slots__ (note that slots are ordered, and you shouldn"t repeat slots that are already in parent classes):

class SlottedWithDict(Child): 
    __slots__ = ("__dict__", "b")

swd = SlottedWithDict()
swd.a = "a"
swd.b = "b"
swd.c = "c"

and

>>> swd.__dict__
{"c": "c"}

Or you don"t even need to declare __slots__ in your subclass, and you will still use slots from the parents, but not restrict the creation of a __dict__:

class NoSlots(Child): pass
ns = NoSlots()
ns.a = "a"
ns.b = "b"

And:

>>> ns.__dict__
{"b": "b"}

However, __slots__ may cause problems for multiple inheritance:

class BaseA(object): 
    __slots__ = ("a",)

class BaseB(object): 
    __slots__ = ("b",)

Because creating a child class from parents with both non-empty slots fails:

>>> class Child(BaseA, BaseB): __slots__ = ()
Traceback (most recent call last):
  File "<pyshell#68>", line 1, in <module>
    class Child(BaseA, BaseB): __slots__ = ()
TypeError: Error when calling the metaclass bases
    multiple bases have instance lay-out conflict

If you run into this problem, You could just remove __slots__ from the parents, or if you have control of the parents, give them empty slots, or refactor to abstractions:

from abc import ABC

class AbstractA(ABC):
    __slots__ = ()

class BaseA(AbstractA): 
    __slots__ = ("a",)

class AbstractB(ABC):
    __slots__ = ()

class BaseB(AbstractB): 
    __slots__ = ("b",)

class Child(AbstractA, AbstractB): 
    __slots__ = ("a", "b")

c = Child() # no problem!

Add "__dict__" to __slots__ to get dynamic assignment:

class Foo(object):
    __slots__ = "bar", "baz", "__dict__"

and now:

>>> foo = Foo()
>>> foo.boink = "boink"

So with "__dict__" in slots we lose some of the size benefits with the upside of having dynamic assignment and still having slots for the names we do expect.

When you inherit from an object that isn"t slotted, you get the same sort of semantics when you use __slots__ - names that are in __slots__ point to slotted values, while any other values are put in the instance"s __dict__.

Avoiding __slots__ because you want to be able to add attributes on the fly is actually not a good reason - just add "__dict__" to your __slots__ if this is required.

You can similarly add __weakref__ to __slots__ explicitly if you need that feature.

Set to empty tuple when subclassing a namedtuple:

The namedtuple builtin make immutable instances that are very lightweight (essentially, the size of tuples) but to get the benefits, you need to do it yourself if you subclass them:

from collections import namedtuple
class MyNT(namedtuple("MyNT", "bar baz")):
    """MyNT is an immutable and lightweight object"""
    __slots__ = ()

usage:

>>> nt = MyNT("bar", "baz")
>>> nt.bar
"bar"
>>> nt.baz
"baz"

And trying to assign an unexpected attribute raises an AttributeError because we have prevented the creation of __dict__:

>>> nt.quux = "quux"
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: "MyNT" object has no attribute "quux"

You can allow __dict__ creation by leaving off __slots__ = (), but you can"t use non-empty __slots__ with subtypes of tuple.

Biggest Caveat: Multiple inheritance

Even when non-empty slots are the same for multiple parents, they cannot be used together:

class Foo(object): 
    __slots__ = "foo", "bar"
class Bar(object):
    __slots__ = "foo", "bar" # alas, would work if empty, i.e. ()

>>> class Baz(Foo, Bar): pass
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: Error when calling the metaclass bases
    multiple bases have instance lay-out conflict

Using an empty __slots__ in the parent seems to provide the most flexibility, allowing the child to choose to prevent or allow (by adding "__dict__" to get dynamic assignment, see section above) the creation of a __dict__:

class Foo(object): __slots__ = ()
class Bar(object): __slots__ = ()
class Baz(Foo, Bar): __slots__ = ("foo", "bar")
b = Baz()
b.foo, b.bar = "foo", "bar"

You don"t have to have slots - so if you add them, and remove them later, it shouldn"t cause any problems.

Going out on a limb here: If you"re composing mixins or using abstract base classes, which aren"t intended to be instantiated, an empty __slots__ in those parents seems to be the best way to go in terms of flexibility for subclassers.

To demonstrate, first, let"s create a class with code we"d like to use under multiple inheritance

class AbstractBase:
    __slots__ = ()
    def __init__(self, a, b):
        self.a = a
        self.b = b
    def __repr__(self):
        return f"{type(self).__name__}({repr(self.a)}, {repr(self.b)})"

We could use the above directly by inheriting and declaring the expected slots:

class Foo(AbstractBase):
    __slots__ = "a", "b"

But we don"t care about that, that"s trivial single inheritance, we need another class we might also inherit from, maybe with a noisy attribute:

class AbstractBaseC:
    __slots__ = ()
    @property
    def c(self):
        print("getting c!")
        return self._c
    @c.setter
    def c(self, arg):
        print("setting c!")
        self._c = arg

Now if both bases had nonempty slots, we couldn"t do the below. (In fact, if we wanted, we could have given AbstractBase nonempty slots a and b, and left them out of the below declaration - leaving them in would be wrong):

class Concretion(AbstractBase, AbstractBaseC):
    __slots__ = "a b _c".split()

And now we have functionality from both via multiple inheritance, and can still deny __dict__ and __weakref__ instantiation:

>>> c = Concretion("a", "b")
>>> c.c = c
setting c!
>>> c.c
getting c!
Concretion("a", "b")
>>> c.d = "d"
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
AttributeError: "Concretion" object has no attribute "d"

Other cases to avoid slots:

• Avoid them when you want to perform __class__ assignment with another class that doesn"t have them (and you can"t add them) unless the slot layouts are identical. (I am very interested in learning who is doing this and why.)
• Avoid them if you want to subclass variable length builtins like long, tuple, or str, and you want to add attributes to them.
• Avoid them if you insist on providing default values via class attributes for instance variables.

You may be able to tease out further caveats from the rest of the __slots__ documentation (the 3.7 dev docs are the most current), which I have made significant recent contributions to.

Critiques of other answers

The current top answers cite outdated information and are quite hand-wavy and miss the mark in some important ways.

Do not "only use __slots__ when instantiating lots of objects"

I quote:

"You would want to use __slots__ if you are going to instantiate a lot (hundreds, thousands) of objects of the same class."

Abstract Base Classes, for example, from the collections module, are not instantiated, yet __slots__ are declared for them.

Why?

If a user wishes to deny __dict__ or __weakref__ creation, those things must not be available in the parent classes.

__slots__ contributes to reusability when creating interfaces or mixins.

It is true that many Python users aren"t writing for reusability, but when you are, having the option to deny unnecessary space usage is valuable.

__slots__ doesn"t break pickling

When pickling a slotted object, you may find it complains with a misleading TypeError:

>>> pickle.loads(pickle.dumps(f))
TypeError: a class that defines __slots__ without defining __getstate__ cannot be pickled

This is actually incorrect. This message comes from the oldest protocol, which is the default. You can select the latest protocol with the -1 argument. In Python 2.7 this would be 2 (which was introduced in 2.3), and in 3.6 it is 4.

>>> pickle.loads(pickle.dumps(f, -1))
<__main__.Foo object at 0x1129C770>

in Python 2.7:

>>> pickle.loads(pickle.dumps(f, 2))
<__main__.Foo object at 0x1129C770>

in Python 3.6

>>> pickle.loads(pickle.dumps(f, 4))
<__main__.Foo object at 0x1129C770>

So I would keep this in mind, as it is a solved problem.

Critique of the (until Oct 2, 2016) accepted answer

The first paragraph is half short explanation, half predictive. Here"s the only part that actually answers the question

The proper use of __slots__ is to save space in objects. Instead of having a dynamic dict that allows adding attributes to objects at anytime, there is a static structure which does not allow additions after creation. This saves the overhead of one dict for every object that uses slots

The second half is wishful thinking, and off the mark:

While this is sometimes a useful optimization, it would be completely unnecessary if the Python interpreter was dynamic enough so that it would only require the dict when there actually were additions to the object.

Python actually does something similar to this, only creating the __dict__ when it is accessed, but creating lots of objects with no data is fairly ridiculous.

The second paragraph oversimplifies and misses actual reasons to avoid __slots__. The below is not a real reason to avoid slots (for actual reasons, see the rest of my answer above.):

They change the behavior of the objects that have slots in a way that can be abused by control freaks and static typing weenies.

It then goes on to discuss other ways of accomplishing that perverse goal with Python, not discussing anything to do with __slots__.

The third paragraph is more wishful thinking. Together it is mostly off-the-mark content that the answerer didn"t even author and contributes to ammunition for critics of the site.

Memory usage evidence

Create some normal objects and slotted objects:

>>> class Foo(object): pass
>>> class Bar(object): __slots__ = ()

Instantiate a million of them:

>>> foos = [Foo() for f in xrange(1000000)]
>>> bars = [Bar() for b in xrange(1000000)]

Inspect with guppy.hpy().heap():

>>> guppy.hpy().heap()
Partition of a set of 2028259 objects. Total size = 99763360 bytes.
 Index  Count   %     Size   % Cumulative  % Kind (class / dict of class)
     0 1000000  49 64000000  64  64000000  64 __main__.Foo
     1     169   0 16281480  16  80281480  80 list
     2 1000000  49 16000000  16  96281480  97 __main__.Bar
     3   12284   1   987472   1  97268952  97 str
...

Access the regular objects and their __dict__ and inspect again:

>>> for f in foos:
...     f.__dict__
>>> guppy.hpy().heap()
Partition of a set of 3028258 objects. Total size = 379763480 bytes.
 Index  Count   %      Size    % Cumulative  % Kind (class / dict of class)
     0 1000000  33 280000000  74 280000000  74 dict of __main__.Foo
     1 1000000  33  64000000  17 344000000  91 __main__.Foo
     2     169   0  16281480   4 360281480  95 list
     3 1000000  33  16000000   4 376281480  99 __main__.Bar
     4   12284   0    987472   0 377268952  99 str
...

This is consistent with the history of Python, from Unifying types and classes in Python 2.2

If you subclass a built-in type, extra space is automatically added to the instances to accomodate __dict__ and __weakrefs__. (The __dict__ is not initialized until you use it though, so you shouldn"t worry about the space occupied by an empty dictionary for each instance you create.) If you don"t need this extra space, you can add the phrase "__slots__ = []" to your class.

Answer #3

os.listdir() - list in the current directory

With listdir in os module you get the files and the folders in the current dir

 import os
 arr = os.listdir()
 print(arr)
 
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

Looking in a directory

arr = os.listdir("c:\files")

glob from glob

with glob you can specify a type of file to list like this

import glob

txtfiles = []
for file in glob.glob("*.txt"):
    txtfiles.append(file)

glob in a list comprehension

mylist = [f for f in glob.glob("*.txt")]

get the full path of only files in the current directory

import os
from os import listdir
from os.path import isfile, join

cwd = os.getcwd()
onlyfiles = [os.path.join(cwd, f) for f in os.listdir(cwd) if 
os.path.isfile(os.path.join(cwd, f))]
print(onlyfiles) 

["G:\getfilesname\getfilesname.py", "G:\getfilesname\example.txt"]

Getting the full path name with os.path.abspath

You get the full path in return

 import os
 files_path = [os.path.abspath(x) for x in os.listdir()]
 print(files_path)
 
 ["F:\documentiapplications.txt", "F:\documenticollections.txt"]

Walk: going through sub directories

os.walk returns the root, the directories list and the files list, that is why I unpacked them in r, d, f in the for loop; it, then, looks for other files and directories in the subfolders of the root and so on until there are no subfolders.

import os

# Getting the current work directory (cwd)
thisdir = os.getcwd()

# r=root, d=directories, f = files
for r, d, f in os.walk(thisdir):
    for file in f:
        if file.endswith(".docx"):
            print(os.path.join(r, file))

os.listdir(): get files in the current directory (Python 2)

In Python 2, if you want the list of the files in the current directory, you have to give the argument as "." or os.getcwd() in the os.listdir method.

 import os
 arr = os.listdir(".")
 print(arr)
 
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

To go up in the directory tree

# Method 1
x = os.listdir("..")

# Method 2
x= os.listdir("/")

Get files: os.listdir() in a particular directory (Python 2 and 3)

 import os
 arr = os.listdir("F:\python")
 print(arr)
 
 >>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]

Get files of a particular subdirectory with os.listdir()

import os

x = os.listdir("./content")

os.walk(".") - current directory

 import os
 arr = next(os.walk("."))[2]
 print(arr)
 
 >>> ["5bs_Turismo1.pdf", "5bs_Turismo1.pptx", "esperienza.txt"]

next(os.walk(".")) and os.path.join("dir", "file")

 import os
 arr = []
 for d,r,f in next(os.walk("F:\_python")):
     for file in f:
         arr.append(os.path.join(r,file))

 for f in arr:
     print(files)

>>> F:\_python\dict_class.py
>>> F:\_python\programmi.txt

next(os.walk("F:\") - get the full path - list comprehension

 [os.path.join(r,file) for r,d,f in next(os.walk("F:\_python")) for file in f]
 
 >>> ["F:\_python\dict_class.py", "F:\_python\programmi.txt"]

os.walk - get full path - all files in sub dirs**

x = [os.path.join(r,file) for r,d,f in os.walk("F:\_python") for file in f]
print(x)

>>> ["F:\_python\dict.py", "F:\_python\progr.txt", "F:\_python\readl.py"]

os.listdir() - get only txt files

 arr_txt = [x for x in os.listdir() if x.endswith(".txt")]
 print(arr_txt)
 
 >>> ["work.txt", "3ebooks.txt"]

Using glob to get the full path of the files

If I should need the absolute path of the files:

from path import path
from glob import glob
x = [path(f).abspath() for f in glob("F:\*.txt")]
for f in x:
    print(f)

>>> F:acquistionline.txt
>>> F:acquisti_2018.txt
>>> F:ootstrap_jquery_ecc.txt

Using os.path.isfile to avoid directories in the list

import os.path
listOfFiles = [f for f in os.listdir() if os.path.isfile(f)]
print(listOfFiles)

>>> ["a simple game.py", "data.txt", "decorator.py"]

Using pathlib from Python 3.4

import pathlib

flist = []
for p in pathlib.Path(".").iterdir():
    if p.is_file():
        print(p)
        flist.append(p)

 >>> error.PNG
 >>> exemaker.bat
 >>> guiprova.mp3
 >>> setup.py
 >>> speak_gui2.py
 >>> thumb.PNG

With list comprehension:

flist = [p for p in pathlib.Path(".").iterdir() if p.is_file()]

Alternatively, use pathlib.Path() instead of pathlib.Path(".")

Use glob method in pathlib.Path()

import pathlib

py = pathlib.Path().glob("*.py")
for file in py:
    print(file)

>>> stack_overflow_list.py
>>> stack_overflow_list_tkinter.py

Get all and only files with os.walk

import os
x = [i[2] for i in os.walk(".")]
y=[]
for t in x:
    for f in t:
        y.append(f)
print(y)

>>> ["append_to_list.py", "data.txt", "data1.txt", "data2.txt", "data_180617", "os_walk.py", "READ2.py", "read_data.py", "somma_defaltdic.py", "substitute_words.py", "sum_data.py", "data.txt", "data1.txt", "data_180617"]

Get only files with next and walk in a directory

 import os
 x = next(os.walk("F://python"))[2]
 print(x)
 
 >>> ["calculator.bat","calculator.py"]

Get only directories with next and walk in a directory

 import os
 next(os.walk("F://python"))[1] # for the current dir use (".")
 
 >>> ["python3","others"]

Get all the subdir names with walk

for r,d,f in os.walk("F:\_python"):
    for dirs in d:
        print(dirs)

>>> .vscode
>>> pyexcel
>>> pyschool.py
>>> subtitles
>>> _metaprogramming
>>> .ipynb_checkpoints

os.scandir() from Python 3.5 and greater

import os
x = [f.name for f in os.scandir() if f.is_file()]
print(x)

>>> ["calculator.bat","calculator.py"]

# Another example with scandir (a little variation from docs.python.org)
# This one is more efficient than os.listdir.
# In this case, it shows the files only in the current directory
# where the script is executed.

import os
with os.scandir() as i:
    for entry in i:
        if entry.is_file():
            print(entry.name)

>>> ebookmaker.py
>>> error.PNG
>>> exemaker.bat
>>> guiprova.mp3
>>> setup.py
>>> speakgui4.py
>>> speak_gui2.py
>>> speak_gui3.py
>>> thumb.PNG

Examples:

Ex. 1: How many files are there in the subdirectories?

In this example, we look for the number of files that are included in all the directory and its subdirectories.

import os

def count(dir, counter=0):
    "returns number of files in dir and subdirs"
    for pack in os.walk(dir):
        for f in pack[2]:
            counter += 1
    return dir + " : " + str(counter) + "files"

print(count("F:\python"))

>>> "F:\python" : 12057 files"

Ex.2: How to copy all files from a directory to another?

A script to make order in your computer finding all files of a type (default: pptx) and copying them in a new folder.

import os
import shutil
from path import path

destination = "F:\file_copied"
# os.makedirs(destination)

def copyfile(dir, filetype="pptx", counter=0):
    "Searches for pptx (or other - pptx is the default) files and copies them"
    for pack in os.walk(dir):
        for f in pack[2]:
            if f.endswith(filetype):
                fullpath = pack[0] + "\" + f
                print(fullpath)
                shutil.copy(fullpath, destination)
                counter += 1
    if counter > 0:
        print("-" * 30)
        print("	==> Found in: `" + dir + "` : " + str(counter) + " files
")

for dir in os.listdir():
    "searches for folders that starts with `_`"
    if dir[0] == "_":
        # copyfile(dir, filetype="pdf")
        copyfile(dir, filetype="txt")


>>> _compiti18Compito Contabilità 1conti.txt
>>> _compiti18Compito Contabilità 1modula4.txt
>>> _compiti18Compito Contabilità 1moduloa4.txt
>>> ------------------------
>>> ==> Found in: `_compiti18` : 3 files

Ex. 3: How to get all the files in a txt file

In case you want to create a txt file with all the file names:

import os
mylist = ""
with open("filelist.txt", "w", encoding="utf-8") as file:
    for eachfile in os.listdir():
        mylist += eachfile + "
"
    file.write(mylist)

Example: txt with all the files of an hard drive

"""
We are going to save a txt file with all the files in your directory.
We will use the function walk()
"""

import os

# see all the methods of os
# print(*dir(os), sep=", ")
listafile = []
percorso = []
with open("lista_file.txt", "w", encoding="utf-8") as testo:
    for root, dirs, files in os.walk("D:\"):
        for file in files:
            listafile.append(file)
            percorso.append(root + "\" + file)
            testo.write(file + "
")
listafile.sort()
print("N. of files", len(listafile))
with open("lista_file_ordinata.txt", "w", encoding="utf-8") as testo_ordinato:
    for file in listafile:
        testo_ordinato.write(file + "
")

with open("percorso.txt", "w", encoding="utf-8") as file_percorso:
    for file in percorso:
        file_percorso.write(file + "
")

os.system("lista_file.txt")
os.system("lista_file_ordinata.txt")
os.system("percorso.txt")

All the file of C: in one text file

This is a shorter version of the previous code. Change the folder where to start finding the files if you need to start from another position. This code generate a 50 mb on text file on my computer with something less then 500.000 lines with files with the complete path.

import os

with open("file.txt", "w", encoding="utf-8") as filewrite:
    for r, d, f in os.walk("C:\"):
        for file in f:
            filewrite.write(f"{r + file}
")

How to write a file with all paths in a folder of a type

With this function you can create a txt file that will have the name of a type of file that you look for (ex. pngfile.txt) with all the full path of all the files of that type. It can be useful sometimes, I think.

import os

def searchfiles(extension=".ttf", folder="H:\"):
    "Create a txt file with all the file of a type"
    with open(extension[1:] + "file.txt", "w", encoding="utf-8") as filewrite:
        for r, d, f in os.walk(folder):
            for file in f:
                if file.endswith(extension):
                    filewrite.write(f"{r + file}
")

# looking for png file (fonts) in the hard disk H:
searchfiles(".png", "H:\")

>>> H:4bs_18Dolphins5.png
>>> H:4bs_18Dolphins6.png
>>> H:4bs_18Dolphins7.png
>>> H:5_18marketing htmlassetsimageslogo2.png
>>> H:7z001.png
>>> H:7z002.png

(New) Find all files and open them with tkinter GUI

I just wanted to add in this 2019 a little app to search for all files in a dir and be able to open them by doubleclicking on the name of the file in the list.

import tkinter as tk
import os

def searchfiles(extension=".txt", folder="H:\"):
    "insert all files in the listbox"
    for r, d, f in os.walk(folder):
        for file in f:
            if file.endswith(extension):
                lb.insert(0, r + "\" + file)

def open_file():
    os.startfile(lb.get(lb.curselection()[0]))

root = tk.Tk()
root.geometry("400x400")
bt = tk.Button(root, text="Search", command=lambda:searchfiles(".png", "H:\"))
bt.pack()
lb = tk.Listbox(root)
lb.pack(fill="both", expand=1)
lb.bind("<Double-Button>", lambda x: open_file())
root.mainloop()

Answer #4

I just used the following which was quite simple. First open a console then cd to where you"ve downloaded your file like some-package.whl and use

pip install some-package.whl

Note: if pip.exe is not recognized, you may find it in the "Scripts" directory from where python has been installed. If pip is not installed, this page can help: How do I install pip on Windows?

Note: for clarification
If you copy the *.whl file to your local drive (ex. C:some-dirsome-file.whl) use the following command line parameters --

pip install C:/some-dir/some-file.whl

Answer #5

Quick Answer:

The simplest way to get row counts per group is by calling .size(), which returns a Series:

df.groupby(["col1","col2"]).size()

Usually you want this result as a DataFrame (instead of a Series) so you can do:

df.groupby(["col1", "col2"]).size().reset_index(name="counts")

If you want to find out how to calculate the row counts and other statistics for each group continue reading below.

Detailed example:

Consider the following example dataframe:

In [2]: df
Out[2]: 
  col1 col2  col3  col4  col5  col6
0    A    B  0.20 -0.61 -0.49  1.49
1    A    B -1.53 -1.01 -0.39  1.82
2    A    B -0.44  0.27  0.72  0.11
3    A    B  0.28 -1.32  0.38  0.18
4    C    D  0.12  0.59  0.81  0.66
5    C    D -0.13 -1.65 -1.64  0.50
6    C    D -1.42 -0.11 -0.18 -0.44
7    E    F -0.00  1.42 -0.26  1.17
8    E    F  0.91 -0.47  1.35 -0.34
9    G    H  1.48 -0.63 -1.14  0.17

First let"s use .size() to get the row counts:

In [3]: df.groupby(["col1", "col2"]).size()
Out[3]: 
col1  col2
A     B       4
C     D       3
E     F       2
G     H       1
dtype: int64

Then let"s use .size().reset_index(name="counts") to get the row counts:

In [4]: df.groupby(["col1", "col2"]).size().reset_index(name="counts")
Out[4]: 
  col1 col2  counts
0    A    B       4
1    C    D       3
2    E    F       2
3    G    H       1

Including results for more statistics

When you want to calculate statistics on grouped data, it usually looks like this:

In [5]: (df
   ...: .groupby(["col1", "col2"])
   ...: .agg({
   ...:     "col3": ["mean", "count"], 
   ...:     "col4": ["median", "min", "count"]
   ...: }))
Out[5]: 
            col4                  col3      
          median   min count      mean count
col1 col2                                   
A    B    -0.810 -1.32     4 -0.372500     4
C    D    -0.110 -1.65     3 -0.476667     3
E    F     0.475 -0.47     2  0.455000     2
G    H    -0.630 -0.63     1  1.480000     1

The result above is a little annoying to deal with because of the nested column labels, and also because row counts are on a per column basis.

To gain more control over the output I usually split the statistics into individual aggregations that I then combine using join. It looks like this:

In [6]: gb = df.groupby(["col1", "col2"])
   ...: counts = gb.size().to_frame(name="counts")
   ...: (counts
   ...:  .join(gb.agg({"col3": "mean"}).rename(columns={"col3": "col3_mean"}))
   ...:  .join(gb.agg({"col4": "median"}).rename(columns={"col4": "col4_median"}))
   ...:  .join(gb.agg({"col4": "min"}).rename(columns={"col4": "col4_min"}))
   ...:  .reset_index()
   ...: )
   ...: 
Out[6]: 
  col1 col2  counts  col3_mean  col4_median  col4_min
0    A    B       4  -0.372500       -0.810     -1.32
1    C    D       3  -0.476667       -0.110     -1.65
2    E    F       2   0.455000        0.475     -0.47
3    G    H       1   1.480000       -0.630     -0.63

Footnotes

The code used to generate the test data is shown below:

In [1]: import numpy as np
   ...: import pandas as pd 
   ...: 
   ...: keys = np.array([
   ...:         ["A", "B"],
   ...:         ["A", "B"],
   ...:         ["A", "B"],
   ...:         ["A", "B"],
   ...:         ["C", "D"],
   ...:         ["C", "D"],
   ...:         ["C", "D"],
   ...:         ["E", "F"],
   ...:         ["E", "F"],
   ...:         ["G", "H"] 
   ...:         ])
   ...: 
   ...: df = pd.DataFrame(
   ...:     np.hstack([keys,np.random.randn(10,4).round(2)]), 
   ...:     columns = ["col1", "col2", "col3", "col4", "col5", "col6"]
   ...: )
   ...: 
   ...: df[["col3", "col4", "col5", "col6"]] = 
   ...:     df[["col3", "col4", "col5", "col6"]].astype(float)
   ...: 

Disclaimer:

If some of the columns that you are aggregating have null values, then you really want to be looking at the group row counts as an independent aggregation for each column. Otherwise you may be misled as to how many records are actually being used to calculate things like the mean because pandas will drop NaN entries in the mean calculation without telling you about it.

Answer #6

Using a for loop, how do I access the loop index, from 1 to 5 in this case?

Use enumerate to get the index with the element as you iterate:

for index, item in enumerate(items):
    print(index, item)

And note that Python"s indexes start at zero, so you would get 0 to 4 with the above. If you want the count, 1 to 5, do this:

count = 0 # in case items is empty and you need it after the loop
for count, item in enumerate(items, start=1):
    print(count, item)

Unidiomatic control flow

What you are asking for is the Pythonic equivalent of the following, which is the algorithm most programmers of lower-level languages would use:

index = 0            # Python"s indexing starts at zero
for item in items:   # Python"s for loops are a "for each" loop 
    print(index, item)
    index += 1

Or in languages that do not have a for-each loop:

index = 0
while index < len(items):
    print(index, items[index])
    index += 1

or sometimes more commonly (but unidiomatically) found in Python:

for index in range(len(items)):
    print(index, items[index])

Use the Enumerate Function

Python"s enumerate function reduces the visual clutter by hiding the accounting for the indexes, and encapsulating the iterable into another iterable (an enumerate object) that yields a two-item tuple of the index and the item that the original iterable would provide. That looks like this:

for index, item in enumerate(items, start=0):   # default is zero
    print(index, item)

This code sample is fairly well the canonical example of the difference between code that is idiomatic of Python and code that is not. Idiomatic code is sophisticated (but not complicated) Python, written in the way that it was intended to be used. Idiomatic code is expected by the designers of the language, which means that usually this code is not just more readable, but also more efficient.

Getting a count

Even if you don"t need indexes as you go, but you need a count of the iterations (sometimes desirable) you can start with 1 and the final number will be your count.

count = 0 # in case items is empty
for count, item in enumerate(items, start=1):   # default is zero
    print(item)

print("there were {0} items printed".format(count))

The count seems to be more what you intend to ask for (as opposed to index) when you said you wanted from 1 to 5.

Breaking it down - a step by step explanation

To break these examples down, say we have a list of items that we want to iterate over with an index:

items = ["a", "b", "c", "d", "e"]

Now we pass this iterable to enumerate, creating an enumerate object:

enumerate_object = enumerate(items) # the enumerate object

We can pull the first item out of this iterable that we would get in a loop with the next function:

iteration = next(enumerate_object) # first iteration from enumerate
print(iteration)

And we see we get a tuple of 0, the first index, and "a", the first item:

(0, "a")

we can use what is referred to as "sequence unpacking" to extract the elements from this two-tuple:

index, item = iteration
#   0,  "a" = (0, "a") # essentially this.

and when we inspect index, we find it refers to the first index, 0, and item refers to the first item, "a".

>>> print(index)
0
>>> print(item)
a

Conclusion

• Python indexes start at zero
• To get these indexes from an iterable as you iterate over it, use the enumerate function
• Using enumerate in the idiomatic way (along with tuple unpacking) creates code that is more readable and maintainable:

So do this:

for index, item in enumerate(items, start=0):   # Python indexes start at zero
    print(index, item)

Answer #7

Getting some sort of modification date in a cross-platform way is easy - just call os.path.getmtime(path) and you"ll get the Unix timestamp of when the file at path was last modified.

Getting file creation dates, on the other hand, is fiddly and platform-dependent, differing even between the three big OSes:

• On Windows, a file"s ctime (documented at https://msdn.microsoft.com/en-us/library/14h5k7ff.aspx) stores its creation date. You can access this in Python through os.path.getctime() or the .st_ctime attribute of the result of a call to os.stat(). This won"t work on Unix, where the ctime is the last time that the file"s attributes or content were changed.
• On Mac, as well as some other Unix-based OSes, you can use the .st_birthtime attribute of the result of a call to os.stat().

• On Linux, this is currently impossible, at least without writing a C extension for Python. Although some file systems commonly used with Linux do store creation dates (for example, ext4 stores them in st_crtime) , the Linux kernel offers no way of accessing them; in particular, the structs it returns from stat() calls in C, as of the latest kernel version, don"t contain any creation date fields. You can also see that the identifier st_crtime doesn"t currently feature anywhere in the Python source. At least if you"re on ext4, the data is attached to the inodes in the file system, but there"s no convenient way of accessing it.

  The next-best thing on Linux is to access the file"s mtime, through either os.path.getmtime() or the .st_mtime attribute of an os.stat() result. This will give you the last time the file"s content was modified, which may be adequate for some use cases.

Putting this all together, cross-platform code should look something like this...

import os
import platform

def creation_date(path_to_file):
    """
    Try to get the date that a file was created, falling back to when it was
    last modified if that isn"t possible.
    See http://stackoverflow.com/a/39501288/1709587 for explanation.
    """
    if platform.system() == "Windows":
        return os.path.getctime(path_to_file)
    else:
        stat = os.stat(path_to_file)
        try:
            return stat.st_birthtime
        except AttributeError:
            # We"re probably on Linux. No easy way to get creation dates here,
            # so we"ll settle for when its content was last modified.
            return stat.st_mtime

Answer #8

I noticed that every now and then I need to Google fopen all over again, just to build a mental image of what the primary differences between the modes are. So, I thought a diagram will be faster to read next time. Maybe someone else will find that helpful too.

Answer #9

I would suggest using the duplicated method on the Pandas Index itself:

df3 = df3[~df3.index.duplicated(keep="first")]

While all the other methods work, .drop_duplicates is by far the least performant for the provided example. Furthermore, while the groupby method is only slightly less performant, I find the duplicated method to be more readable.

Using the sample data provided:

>>> %timeit df3.reset_index().drop_duplicates(subset="index", keep="first").set_index("index")
1000 loops, best of 3: 1.54 ms per loop

>>> %timeit df3.groupby(df3.index).first()
1000 loops, best of 3: 580 µs per loop

>>> %timeit df3[~df3.index.duplicated(keep="first")]
1000 loops, best of 3: 307 µs per loop

Note that you can keep the last element by changing the keep argument to "last".

It should also be noted that this method works with MultiIndex as well (using df1 as specified in Paul"s example):

>>> %timeit df1.groupby(level=df1.index.names).last()
1000 loops, best of 3: 771 µs per loop

>>> %timeit df1[~df1.index.duplicated(keep="last")]
1000 loops, best of 3: 365 µs per loop

Answer #10

Here"s a concise solution which avoids regular expressions and slow in-Python loops:

def principal_period(s):
    i = (s+s).find(s, 1, -1)
    return None if i == -1 else s[:i]

See the Community Wiki answer started by @davidism for benchmark results. In summary,

David Zhang"s solution is the clear winner, outperforming all others by at least 5x for the large example set.

(That answer"s words, not mine.)

This is based on the observation that a string is periodic if and only if it is equal to a nontrivial rotation of itself. Kudos to @AleksiTorhamo for realizing that we can then recover the principal period from the index of the first occurrence of s in (s+s)[1:-1], and for informing me of the optional start and end arguments of Python"s string.find.