How to test if a value is falsy in Python 3

Question

In Javascript, there is == operator to test if a value is falsy:

'' == false // true

In Python, == corresponds to === in Javascript, which is an exact equation (value & type).

So how to find out if a value is Falsy in Python?

Answer 1

You can obtain the truthiness of a value, by using the bool(..) function:

>>> bool('')
False
>>> bool('foo')
True
>>> bool(1)
True
>>> bool(None)
False

In an if statement, the truthiness is calculated implicitly. You can use the not keyword, to invert the truthiness. For example:

>>> not ''
True
>>> not 'foo'
False
>>> not 1
False
>>> not None
True

Answer 2

To get implicit conversion you can just use not - or (for "truthy") just use the variable in place:

if not None:
    print('None')

if not False:
    print('False')

if not '':
    print('empty string')

if not 0:
    print('zero')

if not {}:
    print('empty/zero length container')

if 'hello':
    print('non empty string, truthy test')

Answer 3

What worked was using ternary:

True if '' else False # False

More verbose than in Javascript, but works.

Answer 4

Even tho this question is old, but there is a not not (kinda hacky), but this is the faster than bool(..) and probably the fastest that's possible, you can do it by:

print(not not '')
print(not not 0)
print(not not 'bar')

Output:

False
False
True