Python – Why does “not(True) in [False, True]” return False

comparison-operatorsoperator-precedencepython

If I do this:

>>> False in [False, True]
True

That returns True. Simply because False is in the list.

But if I do:

>>> not(True) in [False, True]
False

That returns False. Whereas not(True) is equal to False:

>>> not(True)
False

Why?

Best Solution

Operator precedence 2.x, 3.x. The precedence of not is lower than that of in. So it is equivalent to:

>>> not ((True) in [False, True])
False

This is what you want:

>>> (not True) in [False, True]
True

As @Ben points out: It's recommended to never write not(True), prefer not True. The former makes it look like a function call, while not is an operator, not a function.