python`list`和`for`返回不同的結果

Question

為什么list並for報告不同的結果，如果我用他們對我的函數生成的值？

from collections import deque

def neighbours(comp0, cand0):
    comp = deque([i for i in comp0])
    cand = deque([i for i in cand0])

    for i in range(len(cand)):
        elem = cand.popleft()
        comp.append(elem)
        yield comp, cand 
        comp.pop()
        cand.append(elem)

    return


>>> n = neighbours([2], [1,4,5])
>>> list(n)
[(deque([2]), deque([1, 4, 5])), (deque([2]), deque([1, 4, 5])), (deque([2]), deque([1, 4, 5]))]
>>> n = neighbours([2], [1,4,5])
>>> for i in n:
...  print(i)
... 
(deque([2, 1]), deque([4, 5]))
(deque([2, 4]), deque([5, 1]))
(deque([2, 5]), deque([1, 4]))
>>> 

Answer 1

您的對象在迭代時會被修改； 當使用for ，您正在打印中間結果，在打印列表時，您將打印最終結果。

如果首先將結果附加到列表，則再次獲得與list()輸出相同的結果：

>>> n = neighbours([2], [1,4,5])
>>> res = []
>>> for i in n:
...     print(i)
...     res.append(i)
... 
(deque([2, 1]), deque([4, 5]))
(deque([2, 4]), deque([5, 1]))
(deque([2, 5]), deque([1, 4]))
>>> res
[(deque([2]), deque([1, 4, 5])), (deque([2]), deque([1, 4, 5])), (deque([2]), deque([1, 4, 5]))]

res中的每個元素都是具有相同兩個雙端隊列對象的元組：

>>> res[0][0] is res[1][0] is res[2][0]
True
>>> res[0][1] is res[1][1] is res[2][1]
True

您可以改為生成每個deque list()副本，從而創建新對象：

>>> def neighbours(comp0, cand0):
...     comp = deque([i for i in comp0])
...     cand = deque([i for i in cand0])
...     for i in range(len(cand)):
...         elem = cand.popleft()
...         comp.append(elem)
...         yield list(comp), list(cand)
...         comp.pop()
...         cand.append(elem)
... 
>>> n = neighbours([2], [1,4,5])
>>> res = []
>>> for i in n:
...     print(i)
...     res.append(i)
... 
([2, 1], [4, 5])
([2, 4], [5, 1])
([2, 5], [1, 4])
>>> res
[([2, 1], [4, 5]), ([2, 4], [5, 1]), ([2, 5], [1, 4])]

Answer 2

哈維爾（Javier）很好地解釋了問題的原因。 現在，如果您希望使用迭代獲得相同的結果，則可以使用：

>>> [i for i in n]
[(deque([2]), deque([1, 4, 5])), (deque([2]), deque([1, 4, 5])), (deque([2]), deque([1, 4, 5]))]