Pandas Series.cummin() is used to find the cumulative minimum of a series. In the cumulative minimum, the returned row is the same length as the input row, and each element is equal to the shorter one between the current element and the previous element.
Syntax: Series.cummin (axis = None, skipna = True)
Parameters:
axis: 0 or `index` for row wise operation and 1 or `columns `for column wise operation
skipna: Skips NaN addition for elements after the very next one if True.Return type: Series
Example # 1:
This example creates a series from a Python list. The list also contains Null, and the skipna parameter remains the default, which is True.
| |
Output:
0 3.0 1 3.0 2 NaN 3 3.0 4 2.0 5 0.0 dtype: float64
Explanation: Cummin — it is a comparison of the current value with the previous value. The first element is always the first of the caller.
3 3 (3 & lt; 4) NaN (Since NaN cannot be compared to integer values) 3 (3 & lt; 7) 2 (2 & lt; 3) 0 (0 & lt; 2)
Example # 2: saving skipna=False
In this example, a series is created in the same way as in the above example. But the skipna parameter remains False. Therefore, NULL values will not be ignored and they will be compared each time they are found.
| |
Exit:
0 12.0 1 4.0 2 4.0 3 NaN 4 NaN 5 NaN 6 NaN 7 NaN dtype: float64
Explanation: Ka to and in the previous example, the minimum of the current and previous values was stored at each position until NaN appeared. Since NaN compared to anything returns NaN and skipna remains False, the cumulative minimum after it appears is NaN due to the comparison of all values to NaN.
Given a list ["foo", "bar", "baz"] and an item in the list "bar", how do I get its index (1) in Python?
In Python, what commands can I use to find:
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 site-packages directory?
How can I find all the files in a directory having the extension .txt in python?
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.
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
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.
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:?
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.
How to iterate over rows in a DataFrame in Pandas?
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):
for loop)DataFrame.apply(): i)  Reductions that can be performed in Cython, ii) Iteration in Python spaceDataFrame.itertuples() and iteritems()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.
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.
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.
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.
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.
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.
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.
In Python, what is the purpose of
__slots__and what are the cases one should avoid this?
The special attribute __slots__ allows you to explicitly state which instance attributes you expect your object instances to have, with the expected results:
The space savings is from
__dict__.__dict__ and __weakref__ creation if parent classes deny them and you declare __slots__.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.
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.
__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.
__slots__: Memory SavingsAnother 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
__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!
"__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.
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.
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"
__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.)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.
The current top answers cite outdated information and are quite hand-wavy and miss the mark in some important ways.
__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 picklingWhen 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.
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.
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.
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)
>>> ["$RECYCLE.BIN", "work.txt", "3ebooks.txt", "documents"]
arr = os.listdir("c:\files")
globfrom glob
with glob you can specify a type of file to list like this
import glob
txtfiles = []
for file in glob.glob("*.txt"):
txtfiles.append(file)
glob in a list comprehensionmylist = [f for f in glob.glob("*.txt")]
import os
from os import listdir
from os.path import isfile, join
cwd = os.getcwd()
onlyfiles = [os.path.join(cwd, f) for f in os.listdir(cwd) if
os.path.isfile(os.path.join(cwd, f))]
print(onlyfiles)
["G:\getfilesname\getfilesname.py", "G:\getfilesname\example.txt"]
Getting the full path name with
os.path.abspath
You get the full path in return
import os
files_path = [os.path.abspath(x) for x in os.listdir()]
print(files_path)
["F:\documentiapplications.txt", "F:\documenticollections.txt"]
Walk: going through sub directories
os.walk returns the root, the directories list and the files list, that is why I unpacked them in r, d, f in the for loop; it, then, looks for other files and directories in the subfolders of the root and so on until there are no subfolders.
import os
# Getting the current work directory (cwd)
thisdir = os.getcwd()
# r=root, d=directories, f = files
for r, d, f in os.walk(thisdir):
for file in f:
if file.endswith(".docx"):
print(os.path.join(r, file))
os.listdir(): get files in the current directory (Python 2)
In Python 2, if you want the list of the files in the current directory, you have to give the argument as "." or os.getcwd() in the os.listdir method.
import os
arr = os.listdir(".")
print(arr)
>>> ["$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("."))andos.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
globto 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.isfileto 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
pathlibfrom 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()
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
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.
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
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
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.
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)
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])
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.
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.
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
So do this:
for index, item in enumerate(items, start=0): # Python indexes start at zero
print(index, item)
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:
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..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
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.
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
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.
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