Associativity of “in” in Python?

Question

I\'m making a Python parser, and this is really confusing me:

>>>  1 in  []  in \'a\'
False

>>> (1 in  []) in \'a\'
TypeError: \'in &         


        

Answer 1

Python does special things with chained comparisons.

The following are evaluated differently:

x > y > z   # in this case, if x > y evaluates to true, then
            # the value of y is being used to compare, again,
            # to z

(x > y) > z # the parenth form, on the other hand, will first
            # evaluate x > y. And, compare the evaluated result
            # with z, which can be "True > z" or "False > z"

In both cases though, if the first comparison is False, the rest of the statement won't be looked at.

For your particular case,

1 in [] in 'a'   # this is false because 1 is not in []

(1 in []) in a   # this gives an error because we are
                 # essentially doing this: False in 'a'

1 in ([] in 'a') # this fails because you cannot do
                 # [] in 'a'

Also to demonstrate the first rule above, these are statements that evaluate to True.

1 in [1,2] in [4,[1,2]] # But "1 in [4,[1,2]]" is False

2 < 4 > 1               # and note "2 < 1" is also not true

Precedence of python operators: http://docs.python.org/reference/expressions.html#summary