Python 表达式求值顺序

Question

I don't see why the following expression evaluates to False :

>>> 1 in [1,2,3] == True
False

in and == has the same precedence and groups left to right, but:

>>> (1 in [1,2,3]) == True
True

and

>>> 1 in ([1,2,3] == True)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: argument of type 'bool' is not iterable

Where is the problem with the parenthesis?

Answer 1

Comparison operators at the same level "chain". You may have seen something such as 10 < x < 25 , which expands to (10 < x) and (x < 25) .

Similarly, your original expression expands to

(1 in [1,2,3]) and ([1,2,3] == True)

The first expression evaluates as True , as you already know. The second expression returns False , giving the result you see.