Weird behaviour of `not` operator with python list

Question

When I'm trying to check whether a list is empty or not using python's not operator it is behaving in a weird manner.

I tried using the not operator with a list to check whether it is empty or not.

>>> a = [] >>> not (a) True >>> not (a) == True True >>> not (a) == False True >>> True == False False

The expected output for not (a) == False should be False.

John Kugelman · Accepted Answer · 2019-09-19 22:35:05Z

11

== has higher precedence than not. not (a) == False is parsed as not (a == False).

edited Sep 19, 2019 at 22:35

answered Sep 19, 2019 at 19:21

John Kugelman

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Comments

Engineero · Accepted Answer · 2019-09-20 13:36:12Z

This is working as expected. Parenthesis added below to clarify how this is being executed:

not (a == True) # True not (a == False) # True

An empty list, a = [], evaluates to False in boolean expressions, but it does not equal False or True. Your expressions in the middle are testing whether it is equal to False or True, so they both evaluate to False, and not False is True.

You can add parenthesis like below to get what you expect:

(not a) == True # True (not a) == False # False

For completeness sake: you can cast to bool to force a Boolean evaluation. bool([]) returns False. So: bool(a) == False and even bool(a) is False are both True.

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