파이썬에서 "in"의 연관성?

파이썬에서 "in"의 연관성?

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저는 파이썬 파서를 만들고 있는데 이것은 정말 혼란 스럽습니다.

>>>  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

연관성 등과 관련하여 Python에서 "in"이 정확히 어떻게 작동합니까?

이 두 표현이 같은 방식으로 작동하지 않는 이유는 무엇입니까?

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1 in [] in 'a'로 평가됩니다 (1 in []) and ([] in 'a').

첫 번째 조건 ( 1 in [])이 False이므로 전체 조건은 다음과 같이 평가됩니다 False. ([] in 'a')실제로 평가되지 않으므로 오류가 발생하지 않습니다.

다음은 명령문 정의입니다.

In [121]: def func():
   .....:     return 1 in [] in 'a'
   .....: 

In [122]: dis.dis(func)
  2           0 LOAD_CONST               1 (1)
              3 BUILD_LIST               0
              6 DUP_TOP             
              7 ROT_THREE           
              8 COMPARE_OP               6 (in)
             11 JUMP_IF_FALSE            8 (to 22)  #if first comparison is wrong 
                                                    #then jump to 22, 
             14 POP_TOP             
             15 LOAD_CONST               2 ('a')
             18 COMPARE_OP               6 (in)     #this is never executed, so no Error
             21 RETURN_VALUE         
        >>   22 ROT_TWO             
             23 POP_TOP             
             24 RETURN_VALUE        

In [150]: def func1():
   .....:     return (1 in  []) in 'a'
   .....: 

In [151]: dis.dis(func1)
  2           0 LOAD_CONST               1 (1)
              3 LOAD_CONST               3 (())
              6 COMPARE_OP               6 (in)   # perform 1 in []
              9 LOAD_CONST               2 ('a')  # now load 'a'
             12 COMPARE_OP               6 (in)   # compare result of (1 in []) with 'a'
                                                  # throws Error coz (False in 'a') is
                                                  # TypeError
             15 RETURN_VALUE   



In [153]: def func2():
   .....:     return 1 in ([] in 'a')
   .....: 

In [154]: dis.dis(func2)
  2           0 LOAD_CONST               1 (1)
              3 BUILD_LIST               0
              6 LOAD_CONST               2 ('a') 
              9 COMPARE_OP               6 (in)  # perform ([] in 'a'), which is 
                                                 # Incorrect, so it throws TypeError
             12 COMPARE_OP               6 (in)  # if no Error then 
                                                 # compare 1 with the result of ([] in 'a')
             15 RETURN_VALUE        

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파이썬은 연결 비교로 특별한 일을합니다.

다음은 다르게 평가됩니다.

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"

두 경우 모두 첫 번째 비교가 False이면 나머지 문은 살펴 보지 않습니다.

특정 경우에

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'

또한 위의 첫 번째 규칙을 보여주기 위해 True로 평가되는 문입니다.

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

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From the documentation:

Comparisons can be chained arbitrarily, e.g., x < y <= z is equivalent to x < y and y <= z, except that y is evaluated only once (but in both cases z is not evaluated at all when x < y is found to be false).

What this means is, that there no associativity in x in y in z!

The following are equivalent:

1 in  []  in 'a'
# <=>
middle = []
#            False          not evaluated
result = (1 in middle) and (middle in 'a')


(1 in  []) in 'a'
# <=>
lhs = (1 in []) # False
result = lhs in 'a' # False in 'a' - TypeError


1 in  ([] in 'a')
# <=>
rhs = ([] in 'a') # TypeError
result = 1 in rhs

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The short answer, since the long one is already given several times here and in excellent ways, is that the boolean expression is short-circuited, this is has stopped evaluation when a change of true in false or vice versa cannot happen by further evaluation.

(see http://en.wikipedia.org/wiki/Short-circuit_evaluation)

It might be a little short (no pun intended) as an answer, but as mentioned, all other explanation is allready done quite well here, but I thought the term deserved to be mentioned.

참고URL : https://stackoverflow.com/questions/12660870/associativity-of-in-in-python