Plot logarithmic axes with matplotlib in python

log | logarithm | StackOverflow

I want to plot a graph with one logarithmic axis using matplotlib.

I"ve been reading the docs, but can"t figure out the syntax. I know that it"s probably something simple like "scale=linear" in the plot arguments, but I can"t seem to get it right

Sample program:

import pylab
import matplotlib.pyplot as plt
a = [pow(10, i) for i in range(10)]
fig = plt.figure()
ax = fig.add_subplot(2, 1, 1)

line, = ax.plot(a, color="blue", lw=2)
pylab.show()

Answer rating: 492

You can use the Axes.set_yscale method. That allows you to change the scale after the Axes object is created. That would also allow you to build a control to let the user pick the scale if you needed to.

The relevant line to add is:

ax.set_yscale("log")

You can use "linear" to switch back to a linear scale. Here"s what your code would look like:

import pylab
import matplotlib.pyplot as plt
a = [pow(10, i) for i in range(10)]
fig = plt.figure()
ax = fig.add_subplot(2, 1, 1)

line, = ax.plot(a, color="blue", lw=2)

ax.set_yscale("log")

pylab.show()

result chart

Answer rating: 325

First of all, it"s not very tidy to mix pylab and pyplot code. What"s more, pyplot style is preferred over using pylab.

Here is a slightly cleaned up code, using only pyplot functions:

from matplotlib import pyplot

a = [ pow(10,i) for i in range(10) ]

pyplot.subplot(2,1,1)
pyplot.plot(a, color="blue", lw=2)
pyplot.yscale("log")
pyplot.show()

The relevant function is pyplot.yscale(). If you use the object-oriented version, replace it by the method Axes.set_yscale(). Remember that you can also change the scale of X axis, using pyplot.xscale() (or Axes.set_xscale()).

Check my question What is the difference between ‘log’ and ‘symlog’? to see a few examples of the graph scales that matplotlib offers.

Answer rating: 60

if you want to change the base of logarithm, just add:

plt.yscale("log",base=2) 

Before Matplotlib 3.3, you would have to use basex/basey as the bases of log





Plot logarithmic axes with matplotlib in python: StackOverflow Questions

Python"s equivalent of && (logical-and) in an if-statement

Question by delete

Here"s my code:

def front_back(a, b):
  # +++your code here+++
  if len(a) % 2 == 0 && len(b) % 2 == 0:
    return a[:(len(a)/2)] + b[:(len(b)/2)] + a[(len(a)/2):] + b[(len(b)/2):] 
  else:
    #todo! Not yet done. :P
  return

I"m getting an error in the IF conditional.
What am I doing wrong?

How do you get the logical xor of two variables in Python?

Question by Zach Hirsch

How do you get the logical xor of two variables in Python?

For example, I have two variables that I expect to be strings. I want to test that only one of them contains a True value (is not None or the empty string):

str1 = raw_input("Enter string one:")
str2 = raw_input("Enter string two:")
if logical_xor(str1, str2):
    print "ok"
else:
    print "bad"

The ^ operator seems to be bitwise, and not defined on all objects:

>>> 1 ^ 1
0
>>> 2 ^ 1
3
>>> "abc" ^ ""
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: unsupported operand type(s) for ^: "str" and "str"

How do I log a Python error with debug information?

I am printing Python exception messages to a log file with logging.error:

import logging
try:
    1/0
except ZeroDivisionError as e:
    logging.error(e)  # ERROR:root:division by zero

Is it possible to print more detailed information about the exception and the code that generated it than just the exception string? Things like line numbers or stack traces would be great.

Making Python loggers output all messages to stdout in addition to log file

Question by user248237

Is there a way to make Python logging using the logging module automatically output things to stdout in addition to the log file where they are supposed to go? For example, I"d like all calls to logger.warning, logger.critical, logger.error to go to their intended places but in addition always be copied to stdout. This is to avoid duplicating messages like:

mylogger.critical("something failed")
print "something failed"

Separation of business logic and data access in django

I am writing a project in Django and I see that 80% of the code is in the file models.py. This code is confusing and, after a certain time, I cease to understand what is really happening.

Here is what bothers me:

  1. I find it ugly that my model level (which was supposed to be responsible only for the work with data from a database) is also sending email, walking on API to other services, etc.
  2. Also, I find it unacceptable to place business logic in the view, because this way it becomes difficult to control. For example, in my application there are at least three ways to create new instances of User, but technically it should create them uniformly.
  3. I do not always notice when the methods and properties of my models become non-deterministic and when they develop side effects.

Here is a simple example. At first, the User model was like this:

class User(db.Models):

    def get_present_name(self):
        return self.name or "Anonymous"

    def activate(self):
        self.status = "activated"
        self.save()

Over time, it turned into this:

class User(db.Models):

    def get_present_name(self): 
        # property became non-deterministic in terms of database
        # data is taken from another service by api
        return remote_api.request_user_name(self.uid) or "Anonymous" 

    def activate(self):
        # method now has a side effect (send message to user)
        self.status = "activated"
        self.save()
        send_mail("Your account is activated!", "…", [self.email])

What I want is to separate entities in my code:

  1. Entities of my database, persistence level: What data does my application keep?
  2. Entities of my application, business logic level: What does my application do?

What are the good practices to implement such an approach that can be applied in Django?

Plot logarithmic axes with matplotlib in python

Question by Jim

I want to plot a graph with one logarithmic axis using matplotlib.

I"ve been reading the docs, but can"t figure out the syntax. I know that it"s probably something simple like "scale=linear" in the plot arguments, but I can"t seem to get it right

Sample program:

import pylab
import matplotlib.pyplot as plt
a = [pow(10, i) for i in range(10)]
fig = plt.figure()
ax = fig.add_subplot(2, 1, 1)

line, = ax.plot(a, color="blue", lw=2)
pylab.show()

logger configuration to log to file and print to stdout

I"m using Python"s logging module to log some debug strings to a file which works pretty well. Now in addition, I"d like to use this module to also print the strings out to stdout. How do I do this? In order to log my strings to a file I use following code:

import logging
import logging.handlers
logger = logging.getLogger("")
logger.setLevel(logging.DEBUG)
handler = logging.handlers.RotatingFileHandler(
    LOGFILE, maxBytes=(1048576*5), backupCount=7
)
formatter = logging.Formatter("%(asctime)s - %(name)s - %(levelname)s - %(message)s")
handler.setFormatter(formatter)
logger.addHandler(handler)

and then call a logger function like

logger.debug("I am written to the file")

Thank you for some help here!

What are logits? What is the difference between softmax and softmax_cross_entropy_with_logits?

In the tensorflow API docs they use a keyword called logits. What is it? A lot of methods are written like:

tf.nn.softmax(logits, name=None)

If logits is just a generic Tensor input, why is it named logits?


Secondly, what is the difference between the following two methods?

tf.nn.softmax(logits, name=None)
tf.nn.softmax_cross_entropy_with_logits(logits, labels, name=None)

I know what tf.nn.softmax does, but not the other. An example would be really helpful.

How can I color Python logging output?

Question by airmind

Some time ago, I saw a Mono application with colored output, presumably because of its log system (because all the messages were standardized).

Now, Python has the logging module, which lets you specify a lot of options to customize output. So, I"m imagining something similar would be possible with Python, but I can’t find out how to do this anywhere.

Is there any way to make the Python logging module output in color?

What I want (for instance) errors in red, debug messages in blue or yellow, and so on.

Of course this would probably require a compatible terminal (most modern terminals are); but I could fallback to the original logging output if color isn"t supported.

Any ideas how I can get colored output with the logging module?

How do I disable log messages from the Requests library?

By default, the Requests python library writes log messages to the console, along the lines of:

Starting new HTTP connection (1): example.com
http://example.com:80 "GET / HTTP/1.1" 200 606

I"m usually not interested in these messages, and would like to disable them. What would be the best way to silence those messages or decrease Requests" verbosity?

Answer #1

The Python 3 range() object doesn"t produce numbers immediately; it is a smart sequence object that produces numbers on demand. All it contains is your start, stop and step values, then as you iterate over the object the next integer is calculated each iteration.

The object also implements the object.__contains__ hook, and calculates if your number is part of its range. Calculating is a (near) constant time operation *. There is never a need to scan through all possible integers in the range.

From the range() object documentation:

The advantage of the range type over a regular list or tuple is that a range object will always take the same (small) amount of memory, no matter the size of the range it represents (as it only stores the start, stop and step values, calculating individual items and subranges as needed).

So at a minimum, your range() object would do:

class my_range:
    def __init__(self, start, stop=None, step=1, /):
        if stop is None:
            start, stop = 0, start
        self.start, self.stop, self.step = start, stop, step
        if step < 0:
            lo, hi, step = stop, start, -step
        else:
            lo, hi = start, stop
        self.length = 0 if lo > hi else ((hi - lo - 1) // step) + 1

    def __iter__(self):
        current = self.start
        if self.step < 0:
            while current > self.stop:
                yield current
                current += self.step
        else:
            while current < self.stop:
                yield current
                current += self.step

    def __len__(self):
        return self.length

    def __getitem__(self, i):
        if i < 0:
            i += self.length
        if 0 <= i < self.length:
            return self.start + i * self.step
        raise IndexError("my_range object index out of range")

    def __contains__(self, num):
        if self.step < 0:
            if not (self.stop < num <= self.start):
                return False
        else:
            if not (self.start <= num < self.stop):
                return False
        return (num - self.start) % self.step == 0

This is still missing several things that a real range() supports (such as the .index() or .count() methods, hashing, equality testing, or slicing), but should give you an idea.

I also simplified the __contains__ implementation to only focus on integer tests; if you give a real range() object a non-integer value (including subclasses of int), a slow scan is initiated to see if there is a match, just as if you use a containment test against a list of all the contained values. This was done to continue to support other numeric types that just happen to support equality testing with integers but are not expected to support integer arithmetic as well. See the original Python issue that implemented the containment test.


* Near constant time because Python integers are unbounded and so math operations also grow in time as N grows, making this a O(log N) operation. Since it’s all executed in optimised C code and Python stores integer values in 30-bit chunks, you’d run out of memory before you saw any performance impact due to the size of the integers involved here.

Answer #2

Since this question was asked in 2010, there has been real simplification in how to do simple multithreading with Python with map and pool.

The code below comes from an article/blog post that you should definitely check out (no affiliation) - Parallelism in one line: A Better Model for Day to Day Threading Tasks. I"ll summarize below - it ends up being just a few lines of code:

from multiprocessing.dummy import Pool as ThreadPool
pool = ThreadPool(4)
results = pool.map(my_function, my_array)

Which is the multithreaded version of:

results = []
for item in my_array:
    results.append(my_function(item))

Description

Map is a cool little function, and the key to easily injecting parallelism into your Python code. For those unfamiliar, map is something lifted from functional languages like Lisp. It is a function which maps another function over a sequence.

Map handles the iteration over the sequence for us, applies the function, and stores all of the results in a handy list at the end.

Enter image description here


Implementation

Parallel versions of the map function are provided by two libraries:multiprocessing, and also its little known, but equally fantastic step child:multiprocessing.dummy.

multiprocessing.dummy is exactly the same as multiprocessing module, but uses threads instead (an important distinction - use multiple processes for CPU-intensive tasks; threads for (and during) I/O):

multiprocessing.dummy replicates the API of multiprocessing, but is no more than a wrapper around the threading module.

import urllib2
from multiprocessing.dummy import Pool as ThreadPool

urls = [
  "http://www.python.org",
  "http://www.python.org/about/",
  "http://www.onlamp.com/pub/a/python/2003/04/17/metaclasses.html",
  "http://www.python.org/doc/",
  "http://www.python.org/download/",
  "http://www.python.org/getit/",
  "http://www.python.org/community/",
  "https://wiki.python.org/moin/",
]

# Make the Pool of workers
pool = ThreadPool(4)

# Open the URLs in their own threads
# and return the results
results = pool.map(urllib2.urlopen, urls)

# Close the pool and wait for the work to finish
pool.close()
pool.join()

And the timing results:

Single thread:   14.4 seconds
       4 Pool:   3.1 seconds
       8 Pool:   1.4 seconds
      13 Pool:   1.3 seconds

Passing multiple arguments (works like this only in Python 3.3 and later):

To pass multiple arrays:

results = pool.starmap(function, zip(list_a, list_b))

Or to pass a constant and an array:

results = pool.starmap(function, zip(itertools.repeat(constant), list_a))

If you are using an earlier version of Python, you can pass multiple arguments via this workaround).

(Thanks to user136036 for the helpful comment.)

Answer #3

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

* 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 #4

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 #5

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. enter image description here

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 #6

This is the behaviour to adopt when the referenced object is deleted. It is not specific to Django; this is an SQL standard. Although Django has its own implementation on top of SQL. (1)

There are seven possible actions to take when such event occurs:

  • CASCADE: When the referenced object is deleted, also delete the objects that have references to it (when you remove a blog post for instance, you might want to delete comments as well). SQL equivalent: CASCADE.
  • PROTECT: Forbid the deletion of the referenced object. To delete it you will have to delete all objects that reference it manually. SQL equivalent: RESTRICT.
  • RESTRICT: (introduced in Django 3.1) Similar behavior as PROTECT that matches SQL"s RESTRICT more accurately. (See django documentation example)
  • SET_NULL: Set the reference to NULL (requires the field to be nullable). For instance, when you delete a User, you might want to keep the comments he posted on blog posts, but say it was posted by an anonymous (or deleted) user. SQL equivalent: SET NULL.
  • SET_DEFAULT: Set the default value. SQL equivalent: SET DEFAULT.
  • SET(...): Set a given value. This one is not part of the SQL standard and is entirely handled by Django.
  • DO_NOTHING: Probably a very bad idea since this would create integrity issues in your database (referencing an object that actually doesn"t exist). SQL equivalent: NO ACTION. (2)

Source: Django documentation

See also the documentation of PostgreSQL for instance.

In most cases, CASCADE is the expected behaviour, but for every ForeignKey, you should always ask yourself what is the expected behaviour in this situation. PROTECT and SET_NULL are often useful. Setting CASCADE where it should not, can potentially delete all of your database in cascade, by simply deleting a single user.


Additional note to clarify cascade direction

It"s funny to notice that the direction of the CASCADE action is not clear to many people. Actually, it"s funny to notice that only the CASCADE action is not clear. I understand the cascade behavior might be confusing, however you must think that it is the same direction as any other action. Thus, if you feel that CASCADE direction is not clear to you, it actually means that on_delete behavior is not clear to you.

In your database, a foreign key is basically represented by an integer field which value is the primary key of the foreign object. Let"s say you have an entry comment_A, which has a foreign key to an entry article_B. If you delete the entry comment_A, everything is fine. article_B used to live without comment_A and don"t bother if it"s deleted. However, if you delete article_B, then comment_A panics! It never lived without article_B and needs it, and it"s part of its attributes (article=article_B, but what is article_B???). This is where on_delete steps in, to determine how to resolve this integrity error, either by saying:

  • "No! Please! Don"t! I can"t live without you!" (which is said PROTECT or RESTRICT in Django/SQL)
  • "All right, if I"m not yours, then I"m nobody"s" (which is said SET_NULL)
  • "Good bye world, I can"t live without article_B" and commit suicide (this is the CASCADE behavior).
  • "It"s OK, I"ve got spare lover, and I"ll reference article_C from now" (SET_DEFAULT, or even SET(...)).
  • "I can"t face reality, and I"ll keep calling your name even if that"s the only thing left to me!" (DO_NOTHING)

I hope it makes cascade direction clearer. :)


Footnotes

(1) Django has its own implementation on top of SQL. And, as mentioned by @JoeMjr2 in the comments below, Django will not create the SQL constraints. If you want the constraints to be ensured by your database (for instance, if your database is used by another application, or if you hang in the database console from time to time), you might want to set the related constraints manually yourself. There is an open ticket to add support for database-level on delete constrains in Django.

(2) Actually, there is one case where DO_NOTHING can be useful: If you want to skip Django"s implementation and implement the constraint yourself at the database-level.

Answer #7

You can also use the option_context, with one or more options:

with pd.option_context("display.max_rows", None, "display.max_columns", None):  # more options can be specified also
    print(df)

This will automatically return the options to their previous values.

If you are working on jupyter-notebook, using display(df) instead of print(df) will use jupyter rich display logic (like so).

Answer #8

The or and and python statements require truth-values. For pandas these are considered ambiguous so you should use "bitwise" | (or) or & (and) operations:

result = result[(result["var"]>0.25) | (result["var"]<-0.25)]

These are overloaded for these kind of datastructures to yield the element-wise or (or and).


Just to add some more explanation to this statement:

The exception is thrown when you want to get the bool of a pandas.Series:

>>> import pandas as pd
>>> x = pd.Series([1])
>>> bool(x)
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

What you hit was a place where the operator implicitly converted the operands to bool (you used or but it also happens for and, if and while):

>>> x or x
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
>>> x and x
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
>>> if x:
...     print("fun")
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().
>>> while x:
...     print("fun")
ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().

Besides these 4 statements there are several python functions that hide some bool calls (like any, all, filter, ...) these are normally not problematic with pandas.Series but for completeness I wanted to mention these.


In your case the exception isn"t really helpful, because it doesn"t mention the right alternatives. For and and or you can use (if you want element-wise comparisons):

  • numpy.logical_or:

    >>> import numpy as np
    >>> np.logical_or(x, y)
    

    or simply the | operator:

    >>> x | y
    
  • numpy.logical_and:

    >>> np.logical_and(x, y)
    

    or simply the & operator:

    >>> x & y
    

If you"re using the operators then make sure you set your parenthesis correctly because of the operator precedence.

There are several logical numpy functions which should work on pandas.Series.


The alternatives mentioned in the Exception are more suited if you encountered it when doing if or while. I"ll shortly explain each of these:

  • If you want to check if your Series is empty:

    >>> x = pd.Series([])
    >>> x.empty
    True
    >>> x = pd.Series([1])
    >>> x.empty
    False
    

    Python normally interprets the length of containers (like list, tuple, ...) as truth-value if it has no explicit boolean interpretation. So if you want the python-like check, you could do: if x.size or if not x.empty instead of if x.

  • If your Series contains one and only one boolean value:

    >>> x = pd.Series([100])
    >>> (x > 50).bool()
    True
    >>> (x < 50).bool()
    False
    
  • If you want to check the first and only item of your Series (like .bool() but works even for not boolean contents):

    >>> x = pd.Series([100])
    >>> x.item()
    100
    
  • If you want to check if all or any item is not-zero, not-empty or not-False:

    >>> x = pd.Series([0, 1, 2])
    >>> x.all()   # because one element is zero
    False
    >>> x.any()   # because one (or more) elements are non-zero
    True
    

Answer #9

If you like ascii art:

  • "VALID" = without padding:

       inputs:         1  2  3  4  5  6  7  8  9  10 11 (12 13)
                      |________________|                dropped
                                     |_________________|
    
  • "SAME" = with zero padding:

                   pad|                                      |pad
       inputs:      0 |1  2  3  4  5  6  7  8  9  10 11 12 13|0  0
                   |________________|
                                  |_________________|
                                                 |________________|
    

In this example:

  • Input width = 13
  • Filter width = 6
  • Stride = 5

Notes:

  • "VALID" only ever drops the right-most columns (or bottom-most rows).
  • "SAME" tries to pad evenly left and right, but if the amount of columns to be added is odd, it will add the extra column to the right, as is the case in this example (the same logic applies vertically: there may be an extra row of zeros at the bottom).

Edit:

About the name:

  • With "SAME" padding, if you use a stride of 1, the layer"s outputs will have the same spatial dimensions as its inputs.
  • With "VALID" padding, there"s no "made-up" padding inputs. The layer only uses valid input data.

Answer #10

⚡️ TL;DR — One line solution.

All you have to do is:

sudo easy_install pip

2019: ⚠️easy_install has been deprecated. Check Method #2 below for preferred installation!

Details:

⚡️ OK, I read the solutions given above, but here"s an EASY solution to install pip.

MacOS comes with Python installed. But to make sure that you have Python installed open the terminal and run the following command.

python --version

If this command returns a version number that means Python exists. Which also means that you already have access to easy_install considering you are using macOS/OSX.

ℹ️ Now, all you have to do is run the following command.

sudo easy_install pip

After that, pip will be installed and you"ll be able to use it for installing other packages.

Let me know if you have any problems installing pip this way.

Cheers!

P.S. I ended up blogging a post about it. QuickTip: How Do I Install pip on macOS or OS X?


✅ UPDATE (Jan 2019): METHOD #2: Two line solution —

easy_install has been deprecated. Please use get-pip.py instead.

First of all download the get-pip file

curl https://bootstrap.pypa.io/get-pip.py -o get-pip.py

Now run this file to install pip

python get-pip.py

That should do it.

Another gif you said? Here ya go!

Plot logarithmic axes with matplotlib in python: StackOverflow Questions

Plot logarithmic axes with matplotlib in python

Question by Jim

I want to plot a graph with one logarithmic axis using matplotlib.

I"ve been reading the docs, but can"t figure out the syntax. I know that it"s probably something simple like "scale=linear" in the plot arguments, but I can"t seem to get it right

Sample program:

import pylab
import matplotlib.pyplot as plt
a = [pow(10, i) for i in range(10)]
fig = plt.figure()
ax = fig.add_subplot(2, 1, 1)

line, = ax.plot(a, color="blue", lw=2)
pylab.show()

How to do exponential and logarithmic curve fitting in Python? I found only polynomial fitting

I have a set of data and I want to compare which line describes it best (polynomials of different orders, exponential or logarithmic).

I use Python and Numpy and for polynomial fitting there is a function polyfit(). But I found no such functions for exponential and logarithmic fitting.

Are there any? Or how to solve it otherwise?

Answer #1

There are lots of things I have seen make a model diverge.

  1. Too high of a learning rate. You can often tell if this is the case if the loss begins to increase and then diverges to infinity.

  2. I am not to familiar with the DNNClassifier but I am guessing it uses the categorical cross entropy cost function. This involves taking the log of the prediction which diverges as the prediction approaches zero. That is why people usually add a small epsilon value to the prediction to prevent this divergence. I am guessing the DNNClassifier probably does this or uses the tensorflow opp for it. Probably not the issue.

  3. Other numerical stability issues can exist such as division by zero where adding the epsilon can help. Another less obvious one if the square root who"s derivative can diverge if not properly simplified when dealing with finite precision numbers. Yet again I doubt this is the issue in the case of the DNNClassifier.

  4. You may have an issue with the input data. Try calling assert not np.any(np.isnan(x)) on the input data to make sure you are not introducing the nan. Also make sure all of the target values are valid. Finally, make sure the data is properly normalized. You probably want to have the pixels in the range [-1, 1] and not [0, 255].

  5. The labels must be in the domain of the loss function, so if using a logarithmic-based loss function all labels must be non-negative (as noted by evan pu and the comments below).

Answer #2

Update:

In Python 3.8, the prod function was added to the math module. See: math.prod().

Older info: Python 3.7 and prior

The function you"re looking for would be called prod() or product() but Python doesn"t have that function. So, you need to write your own (which is easy).

Pronouncement on prod()

Yes, that"s right. Guido rejected the idea for a built-in prod() function because he thought it was rarely needed.

Alternative with reduce()

As you suggested, it is not hard to make your own using reduce() and operator.mul():

from functools import reduce  # Required in Python 3
import operator
def prod(iterable):
    return reduce(operator.mul, iterable, 1)

>>> prod(range(1, 5))
24

Note, in Python 3, the reduce() function was moved to the functools module.

Specific case: Factorials

As a side note, the primary motivating use case for prod() is to compute factorials. We already have support for that in the math module:

>>> import math

>>> math.factorial(10)
3628800

Alternative with logarithms

If your data consists of floats, you can compute a product using sum() with exponents and logarithms:

>>> from math import log, exp

>>> data = [1.2, 1.5, 2.5, 0.9, 14.2, 3.8]
>>> exp(sum(map(log, data)))
218.53799999999993

>>> 1.2 * 1.5 * 2.5 * 0.9 * 14.2 * 3.8
218.53799999999998

Note, the use of log() requires that all the inputs are positive.

Answer #3

Based on a feature mentioned in this answer to another question I have found a very generally applicable solution for placing labels on a bar chart.

Other solutions unfortunately do not work in many cases, because the spacing between label and bar is either given in absolute units of the bars or is scaled by the height of the bar. The former only works for a narrow range of values and the latter gives inconsistent spacing within one plot. Neither works well with logarithmic axes.

The solution I propose works independent of scale (i.e. for small and large numbers) and even correctly places labels for negative values and with logarithmic scales because it uses the visual unit points for offsets.

I have added a negative number to showcase the correct placement of labels in such a case.

The value of the height of each bar is used as a label for it. Other labels can easily be used with Simon"s for rect, label in zip(rects, labels) snippet.

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

# Bring some raw data.
frequencies = [6, -16, 75, 160, 244, 260, 145, 73, 16, 4, 1]

# In my original code I create a series and run on that,
# so for consistency I create a series from the list.
freq_series = pd.Series.from_array(frequencies)

x_labels = [108300.0, 110540.0, 112780.0, 115020.0, 117260.0, 119500.0,
            121740.0, 123980.0, 126220.0, 128460.0, 130700.0]

# Plot the figure.
plt.figure(figsize=(12, 8))
ax = freq_series.plot(kind="bar")
ax.set_title("Amount Frequency")
ax.set_xlabel("Amount ($)")
ax.set_ylabel("Frequency")
ax.set_xticklabels(x_labels)


def add_value_labels(ax, spacing=5):
    """Add labels to the end of each bar in a bar chart.

    Arguments:
        ax (matplotlib.axes.Axes): The matplotlib object containing the axes
            of the plot to annotate.
        spacing (int): The distance between the labels and the bars.
    """

    # For each bar: Place a label
    for rect in ax.patches:
        # Get X and Y placement of label from rect.
        y_value = rect.get_height()
        x_value = rect.get_x() + rect.get_width() / 2

        # Number of points between bar and label. Change to your liking.
        space = spacing
        # Vertical alignment for positive values
        va = "bottom"

        # If value of bar is negative: Place label below bar
        if y_value < 0:
            # Invert space to place label below
            space *= -1
            # Vertically align label at top
            va = "top"

        # Use Y value as label and format number with one decimal place
        label = "{:.1f}".format(y_value)

        # Create annotation
        ax.annotate(
            label,                      # Use `label` as label
            (x_value, y_value),         # Place label at end of the bar
            xytext=(0, space),          # Vertically shift label by `space`
            textcoords="offset points", # Interpret `xytext` as offset in points
            ha="center",                # Horizontally center label
            va=va)                      # Vertically align label differently for
                                        # positive and negative values.


# Call the function above. All the magic happens there.
add_value_labels(ax)

plt.savefig("image.png")

Edit: I have extracted the relevant functionality in a function, as suggested by barnhillec.

This produces the following output:

Bar chart with automatically placed labels on each bar

And with logarithmic scale (and some adjustment to the input data to showcase logarithmic scaling), this is the result:

Bar chart with logarithmic scale with automatically placed labels on each bar

Answer #4

if you want to change the base of logarithm, just add:

plt.yscale("log",base=2) 

Before Matplotlib 3.3, you would have to use basex/basey as the bases of log

Answer #5

Short answer: use not set(a).isdisjoint(b), it"s generally the fastest.

There are four common ways to test if two lists a and b share any items. The first option is to convert both to sets and check their intersection, as such:

bool(set(a) & set(b))

Because sets are stored using a hash table in Python, searching them is O(1) (see here for more information about complexity of operators in Python). Theoretically, this is O(n+m) on average for n and m objects in lists a and b. But 1) it must first create sets out of the lists, which can take a non-negligible amount of time, and 2) it supposes that hashing collisions are sparse among your data.

The second way to do it is using a generator expression performing iteration on the lists, such as:

any(i in a for i in b)

This allows to search in-place, so no new memory is allocated for intermediary variables. It also bails out on the first find. But the in operator is always O(n) on lists (see here).

Another proposed option is an hybridto iterate through one of the list, convert the other one in a set and test for membership on this set, like so:

a = set(a); any(i in a for i in b)

A fourth approach is to take advantage of the isdisjoint() method of the (frozen)sets (see here), for example:

not set(a).isdisjoint(b)

If the elements you search are near the beginning of an array (e.g. it is sorted), the generator expression is favored, as the sets intersection method have to allocate new memory for the intermediary variables:

from timeit import timeit
>>> timeit("bool(set(a) & set(b))", setup="a=list(range(1000));b=list(range(1000))", number=100000)
26.077727576019242
>>> timeit("any(i in a for i in b)", setup="a=list(range(1000));b=list(range(1000))", number=100000)
0.16220548999262974

Here"s a graph of the execution time for this example in function of list size:

Element sharing test execution time when shared at the beginning

Note that both axes are logarithmic. This represents the best case for the generator expression. As can be seen, the isdisjoint() method is better for very small list sizes, whereas the generator expression is better for larger list sizes.

On the other hand, as the search begins with the beginning for the hybrid and generator expression, if the shared element are systematically at the end of the array (or both lists does not share any values), the disjoint and set intersection approaches are then way faster than the generator expression and the hybrid approach.

>>> timeit("any(i in a for i in b)", setup="a=list(range(1000));b=[x+998 for x in range(999,0,-1)]", number=1000))
13.739536046981812
>>> timeit("bool(set(a) & set(b))", setup="a=list(range(1000));b=[x+998 for x in range(999,0,-1)]", number=1000))
0.08102107048034668

Element sharing test execution time when shared at the end

It is interesting to note that the generator expression is way slower for bigger list sizes. This is only for 1000 repetitions, instead of the 100000 for the previous figure. This setup also approximates well when when no elements are shared, and is the best case for the disjoint and set intersection approaches.

Here are two analysis using random numbers (instead of rigging the setup to favor one technique or another):

Element sharing test execution time for randomly generated data with high chance of sharing Element sharing test execution time for randomly generated data with high chance of sharing

High chance of sharing: elements are randomly taken from [1, 2*len(a)]. Low chance of sharing: elements are randomly taken from [1, 1000*len(a)].

Up to now, this analysis supposed both lists are of the same size. In case of two lists of different sizes, for example a is much smaller, isdisjoint() is always faster:

Element sharing test execution time on two differently sized lists when shared at the beginning Element sharing test execution time on two differently sized lists when shared at the end

Make sure that the a list is the smaller, otherwise the performance decreases. In this experiment, the a list size was set constant to 5.

In summary:

  • If the lists are very small (< 10 elements), not set(a).isdisjoint(b) is always the fastest.
  • If the elements in the lists are sorted or have a regular structure that you can take advantage of, the generator expression any(i in a for i in b) is the fastest on large list sizes;
  • Test the set intersection with not set(a).isdisjoint(b), which is always faster than bool(set(a) & set(b)).
  • The hybrid "iterate through list, test on set" a = set(a); any(i in a for i in b) is generally slower than other methods.
  • The generator expression and the hybrid are much slower than the two other approaches when it comes to lists without sharing elements.

In most cases, using the isdisjoint() method is the best approach as the generator expression will take much longer to execute, as it is very inefficient when no elements are shared.

Answer #6

You can do this with the code below, and the code in your question was actually very close to what you needed, all you have to do is call the cmap object you have.

import matplotlib

cmap = matplotlib.cm.get_cmap("Spectral")

rgba = cmap(0.5)
print(rgba) # (0.99807766255210428, 0.99923106502084169, 0.74602077638401709, 1.0)

For values outside of the range [0.0, 1.0] it will return the under and over colour (respectively). This, by default, is the minimum and maximum colour within the range (so 0.0 and 1.0). This default can be changed with cmap.set_under() and cmap.set_over().

For "special" numbers such as np.nan and np.inf the default is to use the 0.0 value, this can be changed using cmap.set_bad() similarly to under and over as above.

Finally it may be necessary for you to normalize your data such that it conforms to the range [0.0, 1.0]. This can be done using matplotlib.colors.Normalize simply as shown in the small example below where the arguments vmin and vmax describe what numbers should be mapped to 0.0 and 1.0 respectively.

import matplotlib

norm = matplotlib.colors.Normalize(vmin=10.0, vmax=20.0)

print(norm(15.0)) # 0.5

A logarithmic normaliser (matplotlib.colors.LogNorm) is also available for data ranges with a large range of values.

(Thanks to both Joe Kington and tcaswell for suggestions on how to improve the answer.)

Answer #7

For fitting y = A + B log x, just fit y against (log x).

>>> x = numpy.array([1, 7, 20, 50, 79])
>>> y = numpy.array([10, 19, 30, 35, 51])
>>> numpy.polyfit(numpy.log(x), y, 1)
array([ 8.46295607,  6.61867463])
# y ≈ 8.46 log(x) + 6.62

For fitting y = AeBx, take the logarithm of both side gives log y = log A + Bx. So fit (log y) against x.

Note that fitting (log y) as if it is linear will emphasize small values of y, causing large deviation for large y. This is because polyfit (linear regression) works by minimizing ∑i (ΔY)2 = ∑i (YiŶi)2. When Yi = log yi, the residues ΔYi = Δ(log yi) ≈ Δyi / |yi|. So even if polyfit makes a very bad decision for large y, the "divide-by-|y|" factor will compensate for it, causing polyfit favors small values.

This could be alleviated by giving each entry a "weight" proportional to y. polyfit supports weighted-least-squares via the w keyword argument.

>>> x = numpy.array([10, 19, 30, 35, 51])
>>> y = numpy.array([1, 7, 20, 50, 79])
>>> numpy.polyfit(x, numpy.log(y), 1)
array([ 0.10502711, -0.40116352])
#    y ≈ exp(-0.401) * exp(0.105 * x) = 0.670 * exp(0.105 * x)
# (^ biased towards small values)
>>> numpy.polyfit(x, numpy.log(y), 1, w=numpy.sqrt(y))
array([ 0.06009446,  1.41648096])
#    y ≈ exp(1.42) * exp(0.0601 * x) = 4.12 * exp(0.0601 * x)
# (^ not so biased)

Note that Excel, LibreOffice and most scientific calculators typically use the unweighted (biased) formula for the exponential regression / trend lines. If you want your results to be compatible with these platforms, do not include the weights even if it provides better results.


Now, if you can use scipy, you could use scipy.optimize.curve_fit to fit any model without transformations.

For y = A + B log x the result is the same as the transformation method:

>>> x = numpy.array([1, 7, 20, 50, 79])
>>> y = numpy.array([10, 19, 30, 35, 51])
>>> scipy.optimize.curve_fit(lambda t,a,b: a+b*numpy.log(t),  x,  y)
(array([ 6.61867467,  8.46295606]), 
 array([[ 28.15948002,  -7.89609542],
        [ -7.89609542,   2.9857172 ]]))
# y ≈ 6.62 + 8.46 log(x)

For y = AeBx, however, we can get a better fit since it computes Δ(log y) directly. But we need to provide an initialize guess so curve_fit can reach the desired local minimum.

>>> x = numpy.array([10, 19, 30, 35, 51])
>>> y = numpy.array([1, 7, 20, 50, 79])
>>> scipy.optimize.curve_fit(lambda t,a,b: a*numpy.exp(b*t),  x,  y)
(array([  5.60728326e-21,   9.99993501e-01]),
 array([[  4.14809412e-27,  -1.45078961e-08],
        [ -1.45078961e-08,   5.07411462e+10]]))
# oops, definitely wrong.
>>> scipy.optimize.curve_fit(lambda t,a,b: a*numpy.exp(b*t),  x,  y,  p0=(4, 0.1))
(array([ 4.88003249,  0.05531256]),
 array([[  1.01261314e+01,  -4.31940132e-02],
        [ -4.31940132e-02,   1.91188656e-04]]))
# y ≈ 4.88 exp(0.0553 x). much better.

comparison of exponential regression

Answer #8

It"s good to know that

log_b(a) = log(a)/log(b)

but also know that math.log takes an optional second argument which allows you to specify the base:

In [22]: import math

In [23]: math.log?
Type:       builtin_function_or_method
Base Class: <type "builtin_function_or_method">
String Form:    <built-in function log>
Namespace:  Interactive
Docstring:
    log(x[, base]) -> the logarithm of x to the given base.
    If the base not specified, returns the natural logarithm (base e) of x.


In [25]: math.log(8,2)
Out[25]: 3.0

Answer #9

A naive algorithm won"t give good results when applied to real-world data. Here is a 20-line algorithm that exploits relative word frequency to give accurate results for real-word text.

(If you want an answer to your original question which does not use word frequency, you need to refine what exactly is meant by "longest word": is it better to have a 20-letter word and ten 3-letter words, or is it better to have five 10-letter words? Once you settle on a precise definition, you just have to change the line defining wordcost to reflect the intended meaning.)

The idea

The best way to proceed is to model the distribution of the output. A good first approximation is to assume all words are independently distributed. Then you only need to know the relative frequency of all words. It is reasonable to assume that they follow Zipf"s law, that is the word with rank n in the list of words has probability roughly 1/(n log N) where N is the number of words in the dictionary.

Once you have fixed the model, you can use dynamic programming to infer the position of the spaces. The most likely sentence is the one that maximizes the product of the probability of each individual word, and it"s easy to compute it with dynamic programming. Instead of directly using the probability we use a cost defined as the logarithm of the inverse of the probability to avoid overflows.

The code

from math import log

# Build a cost dictionary, assuming Zipf"s law and cost = -math.log(probability).
words = open("words-by-frequency.txt").read().split()
wordcost = dict((k, log((i+1)*log(len(words)))) for i,k in enumerate(words))
maxword = max(len(x) for x in words)

def infer_spaces(s):
    """Uses dynamic programming to infer the location of spaces in a string
    without spaces."""

    # Find the best match for the i first characters, assuming cost has
    # been built for the i-1 first characters.
    # Returns a pair (match_cost, match_length).
    def best_match(i):
        candidates = enumerate(reversed(cost[max(0, i-maxword):i]))
        return min((c + wordcost.get(s[i-k-1:i], 9e999), k+1) for k,c in candidates)

    # Build the cost array.
    cost = [0]
    for i in range(1,len(s)+1):
        c,k = best_match(i)
        cost.append(c)

    # Backtrack to recover the minimal-cost string.
    out = []
    i = len(s)
    while i>0:
        c,k = best_match(i)
        assert c == cost[i]
        out.append(s[i-k:i])
        i -= k

    return " ".join(reversed(out))

which you can use with

s = "thumbgreenappleactiveassignmentweeklymetaphor"
print(infer_spaces(s))

The results

I am using this quick-and-dirty 125k-word dictionary I put together from a small subset of Wikipedia.

Before: thumbgreenappleactiveassignmentweeklymetaphor.
After: thumb green apple active assignment weekly metaphor.

Before: thereismassesoftextinformationofpeoplescommentswhichisparsedfromhtmlbuttherearen odelimitedcharactersinthemforexamplethumbgreenappleactiveassignmentweeklymetapho rapparentlytherearethumbgreenappleetcinthestringialsohavealargedictionarytoquery whetherthewordisreasonablesowhatsthefastestwayofextractionthxalot.

After: there is masses of text information of peoples comments which is parsed from html but there are no delimited characters in them for example thumb green apple active assignment weekly metaphor apparently there are thumb green apple etc in the string i also have a large dictionary to query whether the word is reasonable so what s the fastest way of extraction thx a lot.

Before: itwasadarkandstormynighttherainfellintorrentsexceptatoccasionalintervalswhenitwascheckedbyaviolentgustofwindwhichsweptupthestreetsforitisinlondonthatoursceneliesrattlingalongthehousetopsandfiercelyagitatingthescantyflameofthelampsthatstruggledagainstthedarkness.

After: it was a dark and stormy night the rain fell in torrents except at occasional intervals when it was checked by a violent gust of wind which swept up the streets for it is in london that our scene lies rattling along the housetops and fiercely agitating the scanty flame of the lamps that struggled against the darkness.

As you can see it is essentially flawless. The most important part is to make sure your word list was trained to a corpus similar to what you will actually encounter, otherwise the results will be very bad.


Optimization

The implementation consumes a linear amount of time and memory, so it is reasonably efficient. If you need further speedups, you can build a suffix tree from the word list to reduce the size of the set of candidates.

If you need to process a very large consecutive string it would be reasonable to split the string to avoid excessive memory usage. For example you could process the text in blocks of 10000 characters plus a margin of 1000 characters on either side to avoid boundary effects. This will keep memory usage to a minimum and will have almost certainly no effect on the quality.

Answer #10

I finally found some time to do some experiments in order to understand the difference between them. Here"s what I discovered:

  • log only allows positive values, and lets you choose how to handle negative ones (mask or clip).
  • symlog means symmetrical log, and allows positive and negative values.
  • symlog allows to set a range around zero within the plot will be linear instead of logarithmic.

I think everything will get a lot easier to understand with graphics and examples, so let"s try them:

import numpy
from matplotlib import pyplot

# Enable interactive mode
pyplot.ion()

# Draw the grid lines
pyplot.grid(True)

# Numbers from -50 to 50, with 0.1 as step
xdomain = numpy.arange(-50,50, 0.1)

# Plots a simple linear function "f(x) = x"
pyplot.plot(xdomain, xdomain)
# Plots "sin(x)"
pyplot.plot(xdomain, numpy.sin(xdomain))

# "linear" is the default mode, so this next line is redundant:
pyplot.xscale("linear")

A graph using 'linear' scaling

# How to treat negative values?
# "mask" will treat negative values as invalid
# "mask" is the default, so the next two lines are equivalent
pyplot.xscale("log")
pyplot.xscale("log", nonposx="mask")

A graph using 'log' scaling and nonposx='mask'

# "clip" will map all negative values a very small positive one
pyplot.xscale("log", nonposx="clip")

A graph using 'log' scaling and nonposx='clip'

# "symlog" scaling, however, handles negative values nicely
pyplot.xscale("symlog")

A graph using 'symlog' scaling

# And you can even set a linear range around zero
pyplot.xscale("symlog", linthreshx=20)

A graph using 'symlog' scaling, but linear within (-20,20)

Just for completeness, I"ve used the following code to save each figure:

# Default dpi is 80
pyplot.savefig("matplotlib_xscale_linear.png", dpi=50, bbox_inches="tight")

Remember you can change the figure size using:

fig = pyplot.gcf()
fig.set_size_inches([4., 3.])
# Default size: [8., 6.]

(If you are unsure about me answering my own question, read this)

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