python`list`和`for`返回不同的结果

Question

为什么list和for在函数生成的值上使用它们时会报告不同的结果？你知道吗

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

迭代时对象会被修改；使用for时打印中间结果，打印列表时打印最终结果。你知道吗

如果先将结果附加到列表中，则会再次得到与list()输出相同的结果：

>>> 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中的每个元素都是一个元组，具有两个相同的deque对象：

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

您可以生成每个deque的list()副本，从而创建新的对象：

>>> 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

哈维尔已经很好地解释了这个问题的原因。现在，如果希望使用迭代得到相同的结果，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]))]