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

Question

If I do this:

>>> False in [False, True]
True

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

Answer 1

Alongside the other answers that mentioned the precedence of not is lower than in, actually your statement is equivalent to :

not (True in [False, True])

But note that if you don't separate your condition from the other ones, python will use 2 roles (precedence or chaining) in order to separate that, and in this case python used precedence. Also, note that if you want to separate a condition you need to put all the condition in parenthesis not just the object or value :

(not True) in [False, True]

---

But as mentioned, there is another modification by python on operators that is chaining:

Based on python documentation :

Note that comparisons, membership tests, and identity tests, all have the same precedence and have a left-to-right chaining feature as described in the Comparisons section.

For example the result of following statement is False:

>>> True == False in [False, True]
False

Because python will chain the statements like following :

(True == False) and (False in [False, True])

Which exactly is False and True that is False.

You can assume that the central object will be shared between 2 operations and other objects (False in this case).

And note that its also true for all Comparisons, including membership tests and identity tests operations which are following operands :

in, not in, is, is not, <, <=, >, >=, !=, ==

Example :

>>> 1 in [1,2] == True
False

Another famous example is number range :

7

which is equal to :

7