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Simplifying "a == True:" to "a" - is it a good idea?

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:

Python list indexing with bool values [duplicate]

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.

Python :: bool(float('0.0')) -> returns False ? Why? [duplicate]

This question already has answers here:
Python evaluates 0 as False
(5 answers)
Closed 5 years ago.
bool(float('0.0')) -> returns False? Why?
Whereas I can easily type case float('0.0') as 0.0 without any exceptions.

Generally in many programming languages numerical value of 0 is treated as convertible to logical False. Conversely, you may try bool(float('0.1')) and see it's evaluated to True (as 0.1 != 0).

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.2.5
All other values are considered true -- so objects of many types are always true.

Python's Evaluation of If-Else Conditional Statements

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.

Why does using an integer value as a 'while' loop condition work in Python?

The following code will output infinite lines of "test".
foo = 5
while foo:
print("bar")
The other day I came across an answer here about digit sums. This was the code shown in the answer:
def digit_sum(t):
s = 0
while t:
s += t % 10
t //= 10
return s
The part I'm focusing on is the "while t:" part. How and why does this work?

The while condition tests for truth. Any non-zero numeric value is considered true. See the Truth Value Testing section in the Python documentation:
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.
Bold emphasis mine.
In your sample while loop, t trends to 0 (integer division by 10), so eventually while t: ends because t is considered false.

You already got useful answers, but I just wanted to answer your question in a way that can be easily understood by someone who is a beginner in Python.
You could rewrite your code as:
def digit_sum(t):
s = 0
while t!=0:
s += t % 10
t = t//10
return s
'while t' is equivalent to 'while t!=0', meaning that the loop will end when t is equal to 0.
In your for loop, 't //= 10' is equivalent to 't = t // 10' ('//' is a floor division operator and returns an integer). So the value of t becomes smaller each time the loop is executed, until it eventually reaches the value of 0. At this point, the 'while t' condition is False and the loop ends.