python operator precedence of in and comparison

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Question :

python operator precedence of in and comparison

The following comparisons produce True:

>>> '1' in '11'
True
>>> ('1' in '11') == True
True

And with the parentheses the other way, I get a TypeError:

>>> '1' in ('11' == True)
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
TypeError: argument of type 'bool' is not iterable

So how do I get False with no parentheses?

>>> '1' in '11' == True
False
Asked By: nobody

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Answer #1:

The Python manual says in and == are of equal precedence. Thus, they’re evaluated from left to right by default, but there’s also chaining to consider. The expression you put above ('1' in '11' == True) is actually being evaluated as…

('1' in '11') and ('11' == True)

which of course is False. If you don’t know what chaining is, it’s what allows you to do something like…

if 0 < a < 1:

in Python, and have that mean what you expect (“a is greater than 0 but less than 1”).

Answered By: Amber

Answer #2:

It has nothing to do with precedence. In Python relational operators chain, and containment is considered a relational operator. Therefore:

'1' in '11' == True

is the same as:

('1' in '11') and ('11' == True)

which is false since True is not equal to “11”.

Answer #3:

Chaining allows you to write x < y < z, and mean x < y and y < z. Look at this interaction:

>>> (False == True) == False
True
>>> False == (True == False)
True
>>> False == True == False
False
>>>

So in your example, '1' in '11' == True is equivalent to ('1' in '11') and ('11' == True)

Answered By: Roshan Mathews

Answer #4:

What happen here?

'1' in '11' == True  ==>  False

is the same that:

'1' in ('11' == True)

but

('11' == True) ==> False

and

'1' in False

is not defined.

Answered By: Andersson Melo

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