PEP8 suggests the following code should be simplified.
The original
if a == True:
The suggestion
if a:
However, these two are not the same. I figured out when I followed the PEP8 recommendation. Try with the following code
import numpy as np
a = np.nan
if a == True:
print('a is True')
else:
print('a is not True')
if a:
print('a is True')
else:
print('a is not True')
And you will figure out that the first tells a is not true (correctly) while the second one incorrectly tells a is true.
a is not True
a is True
What is the point of this misleading suggestion?
You are misreading the PEP8 style guide. Here is the relevant part (emphasis mine):
Don't compare boolean values to True or False using ==:
# Correct:
if greeting:
# Wrong:
if greeting == True:
Since np.nan is not a boolean value, this advice does not apply.
Note that if you are comparing a numeric value to True, then you are normally doing something wrong in the first place. The numeric values 1 and 1.0 are both equal to True, so if you have a variable which could be either numeric or boolean, this test may give you unexpected results. It is also generally an anti-pattern to have a variable which could be either a boolean or something other than a boolean.
First off, np.nan is works the same way as float('nan').
import numpy as np
print(type(np.nan)) # <class 'float'>
Python normally says:
By default, an object is considered true unless its class defines
either a __bool__() method that returns False or a __len__() method
that returns zero, when called with the object.
Then for built-in numeric types, it says any zeros are considered False:
zero of any numeric type: 0, 0.0, 0j, Decimal(0), Fraction(0, 1)
As I bolded the only float type which is False, any other float numbers are considered True.
so :
print(bool(float('nan'))) # True
Numpy also acts like how python does.
When you say if obj: python tries to get the truth value of the obj by the help of bool() which indeed looks at __bool__ and __len__ special methods. (__bool__ has higher priority if implemented).
I would suggest to use the explicit conditional. The second option will always give you True if a!=0, besides, that type of conditionals are confusing when you didn't write the code.
This means that the variable a has a value equal to True or not.
if a == True:
But does this variable a have a value or not.
if a:
Related
Is it guaranteed that False == 0 and True == 1, in Python (assuming that they are not reassigned by the user)? For instance, is it in any way guaranteed that the following code will always produce the same results, whatever the version of Python (both existing and, likely, future ones)?
0 == False # True
1 == True # True
['zero', 'one'][False] # is 'zero'
Any reference to the official documentation would be much appreciated!
Edit: As noted in many answers, bool inherits from int. The question can therefore be recast as: "Does the documentation officially say that programmers can rely on booleans inheriting from integers, with the values 0 and 1?". This question is relevant for writing robust code that won't fail because of implementation details!
In Python 2.x this is not guaranteed as it is possible for True and False to be reassigned. However, even if this happens, boolean True and boolean False are still properly returned for comparisons.
In Python 3.x True and False are keywords and will always be equal to 1 and 0.
Under normal circumstances in Python 2, and always in Python 3:
False object is of type bool which is a subclass of int:
object
|
int
|
bool
It is the only reason why in your example, ['zero', 'one'][False] does work. It would not work with an object which is not a subclass of integer, because list indexing only works with integers, or objects that define a __index__ method (thanks mark-dickinson).
Edit:
It is true of the current python version, and of that of Python 3. The docs for python 2 and the docs for Python 3 both say:
There are two types of integers: [...] Integers (int) [...] Booleans (bool)
and in the boolean subsection:
Booleans: These represent the truth values False and True [...] Boolean values behave like the values 0 and 1, respectively, in almost all contexts, the exception being that when converted to a string, the strings "False" or "True" are returned, respectively.
There is also, for Python 2:
In numeric contexts (for example when used as the argument to an arithmetic operator), they [False and True] behave like the integers 0 and 1, respectively.
So booleans are explicitly considered as integers in Python 2 and 3.
So you're safe until Python 4 comes along. ;-)
Here's the PEP discussing the new bool type in Python 2.3: http://www.python.org/dev/peps/pep-0285/.
When converting a bool to an int, the integer value is always 0 or 1, but when converting an int to a bool, the boolean value is True for all integers except 0.
>>> int(False)
0
>>> int(True)
1
>>> bool(5)
True
>>> bool(-5)
True
>>> bool(0)
False
In Python 2.x, it is not guaranteed at all:
>>> False = 5
>>> 0 == False
False
So it could change. In Python 3.x, True, False, and None are reserved words, so the above code would not work.
In general, with booleans you should assume that while False will always have an integer value of 0 (so long as you don't change it, as above), True could have any other value. I wouldn't necessarily rely on any guarantee that True==1, but on Python 3.x, this will always be the case, no matter what.
In my url.py file in Django, I have the following line:
if bool(settings.DEBUG) is True:
This works well because settings.DEBUG == 1 therefore it equals True after the bool() function. However, I needed to add the bool() to make it work. The original line was not working and the original line was:
if settings.DEBUG is True:
This always evaluates to False, even when settings.DEBUG == 1.
Why is the value 1 evaluating to False when I do not cast it explicitly to bool?
In other words, why doesn't "1 is True" evaluate true?
Note that I am using python 3 in case that matters to the answer.
At first glance, this can be confusing as of course, the value of '1' should equal true right? However what if this int value equals '99' should this be true also?
By comparing an int with a Boolean value within the python compiler it is like comparing apples with oranges which isn't possible therefore defaulting to 'false'.
By wrapping your int value in a bool(function) you are declaring to the python compiler treat this int value as a boolean (0 == False, 1 == True) therefore it works as you intended.
Hope this helps :)
I recently encountered an example of an if-else conditional statement and could not understand the rationale behind its output. The following are the statements:
if 0:
1
else:
2
Output: 2
I tried different integers in 0's place, and received 1 each time. Is this because the zero in the if condition represents False? But then why do integers other than 1 still satisfy the if condition?
Thanks!
Edit: Thank you for all your answers. I now understand that any integer except 0 in the 'if' statement will make the statement True by default, resulting in an output of 1, in this case.
Python will always attempt to determine the "truthiness" of a given value used in a boolean context. In Python any numerical value of 0 (or 0.0) is considered false, and string, dictionary, list, or other iterable (or other class that can report its length) is false if it's empty or has length of 0. Also, None and boolean False are considered false.
Other values are considered true.
More details: https://docs.python.org/2.4/lib/truth.html.
In Python, bool is a subtype of int. False has the value 0, while other non-zero integers have the subtype bool with the value True.
To see this for yourself try this: False == 0
And to see the subtypes of int try this: int.__subclasses__()
1 is considered True while 0 is False,just like in binary.
Any non-zero numeric value is evaluated as True in a conditional statement.
bool type is just a subtype of int in Python, with 1 == True and 0 == False.
I have come across this piece of code:
counts = 128 * [0]
# some other steps here that modify the variable counts...
# in the end we have something like counts=[12, 583, 0, etc. , etc., 384]
for i in range(len(counts)):
if counts[i]:
outfile.write("%-12s%d\n" % (display(i), counts[i]))
Regarding the if statement, I understand that it is introduced so that it skips the instruction when counts[i]==0. That's because 0 is equivalent to False. But on the other side, I thought that any other integer (apart from 1) was not equivalent to True. In fact, if I type 3 == True on the shell, I get False as an answer. So, is there any particular reason why if <some integer> is equivalent to if True?
3 == True is comparing 3-as-an-integer to True-as-a-boolean.
if 3: is coercing 3 to a boolean.
You'll see that bool(3) == True is true.
Quoting docs:
Any object can be tested for truth value, for use in an if or while
condition or as operand of the Boolean operations below. The following
values are considered false:
None
False
zero of any numeric type, for example, 0, 0L, 0.0, 0j.
any empty sequence, for example, '', (), [].
any empty mapping, for example, {}.
instances of user-defined classes, if the class defines a __nonzero__() or __len__() method, when that method returns the
integer zero or bool value False.
All other values are considered true — so objects of many types are always true.
Simple in your shell try bool(3)(which is True) and bool(0)(which is False).
bool of zero is False and bool of non-zero is True.
When casting a NumPy Not-a-Number value as a boolean, it becomes True, e.g. as follows.
>>> import numpy as np
>>> bool(np.nan)
True
This is the exact opposite to what I would intuitively expect. Is there a sound principle underlying this behaviour?
(I suspect there might be as the same behaviour seems to occur in Octave.)
This is in no way NumPy-specific, but is consistent with how Python treats NaNs:
In [1]: bool(float('nan'))
Out[1]: True
The rules are spelled out in the documentation.
I think it could be reasonably argued that the truth value of NaN should be False. However, this is not how the language works right now.
Python truth-value testing states that the following values are considered False:
zero of any numeric type, for example, 0, 0L, 0.0, 0j.
Numpy probably chose to stick with this behaviour and prevent NaN from evaluating to False in a boolean context. Note however that you can use numpy.isnan to test for NaN.
0.0 is the only falsy float value because that's what the language designers decided would be most useful. Numpy simply follows along. (It would be weird to have bool(np.nan) be False when bool(float('nan')) is True).
I think it is probably because that's how things work with integers. Admittedly, integers have no NaN or inf types of values, but I suppose that special cases aren't special enough to break the rules.
Numpy follows the python standard for truth testing here, any numeric type evaluates to False if and only if its numerical value is zero.
Note that truth testing with NaN values can be unintuitive in other ways as well (e.g., nan == nan evaluates to False).
This is usually needed when testing variables for np.nan. While np.nan is not np.nan fails (evaluates to False), the np.isnan function works (evaluates to True, so can be used to test for np.nan):
np.isnan(np.nan)
True