1 in [] in 'a' is evaluated as (1 in []) and ([] in 'a').ยน
Since the first condition (1 in []) is False, the whole condition is evaluated as False; ([] in 'a') is never actually evaluated, so no error is raised.
We can see how Python executes each statement using the dis module:
>>> from dis import dis
>>> dis("1 in [] in 'a'")
1 0 LOAD_CONST 0 (1)
2 BUILD_LIST 0
4 DUP_TOP
6 ROT_THREE
8 CONTAINS_OP 0 # `in` is the contains operator
10 JUMP_IF_FALSE_OR_POP 18 # skip to 18 if the first
# comparison is false
12 LOAD_CONST 1 ('a') # 12-16 are never executed
14 CONTAINS_OP 0 # so no error here (14)
16 RETURN_VALUE
>> 18 ROT_TWO
20 POP_TOP
22 RETURN_VALUE
>>> dis("(1 in []) in 'a'")
1 0 LOAD_CONST 0 (1)
2 LOAD_CONST 1 (())
4 CONTAINS_OP 0 # perform 1 in []
6 LOAD_CONST 2 ('a') # now load 'a'
8 CONTAINS_OP 0 # check if result of (1 in []) is in 'a'
# throws Error because (False in 'a')
# is a TypeError
10 RETURN_VALUE
>>> dis("1 in ([] in 'a')")
1 0 LOAD_CONST 0 (1)
2 BUILD_LIST 0
4 LOAD_CONST 1 ('a')
6 CONTAINS_OP 0 # perform ([] in 'a'), which is
# incorrect, so it throws a TypeError
8 CONTAINS_OP 0 # if no Error then this would
# check if 1 is in the result of ([] in 'a')
10 RETURN_VALUE
- Except that
[]is only evaluated once. This doesn’t matter in this example but if you (for example) replaced[]with a function that returned a list, that function would only be called once (at most). The documentation explains also this.