Python 中“in”的结合性？

我正在制作一个Python解析器，这是really让我困惑：

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

>>> (1 in []) in 'a'
TypeError: 'in <string>' requires string as left operand, not bool

>>> 1 in ([] in 'a')
TypeError: 'in <string>' requires string as left operand, not list

究竟如何in在Python中工作，关于关联性等？

为什么这些表达式中没有两个表现相同？

1 in [] in 'a'被评估为(1 in []) and ([] in 'a').¹

由于第一个条件 (1 in []) is False，整个条件评估为False; ([] in 'a')从未实际评估过，因此不会引发错误。

我们可以看到Python如何使用以下命令执行每条语句dis module https://docs.python.org/3/library/dis.html:

>>> 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

1. 除了那个[]仅评估一次。这在本例中并不重要，但如果您（例如）替换[]对于返回列表的函数，该函数只会被调用一次（最多）。文档 https://docs.python.org/3/reference/expressions.html#comparisons也解释了这一点。