[英]python `list` and `for` return different results
Why do list
and for
report different result if I use them on the values generated by my function? 为什么
list
并for
报告不同的结果,如果我用他们对我的函数生成的值?
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]))
>>>
Your objects are modified as you iterate; 您的对象在迭代时会被修改; you are printing intermediate results when using
for
, final results when printing the list. 当使用
for
,您正在打印中间结果,在打印列表时,您将打印最终结果。
If you append the results to a list first, you get the same results as your list()
output again: 如果首先将结果附加到列表,则再次获得与
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]))]
Each element in res
is a tuple with the same two deque objects : res
中的每个元素都是具有相同两个双端队列对象的元组:
>>> res[0][0] is res[1][0] is res[2][0]
True
>>> res[0][1] is res[1][1] is res[2][1]
True
You could yield list()
copies of each deque
instead, and thus create new objects: 您可以改为生成每个
deque
list()
副本,从而创建新对象:
>>> 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])]
The reason for the problem has been well explained by Javier . 哈维尔(Javier)很好地解释了问题的原因。 Now if you want same results using iteration you can use:
现在,如果您希望使用迭代获得相同的结果,则可以使用:
>>> [i for i in n]
[(deque([2]), deque([1, 4, 5])), (deque([2]), deque([1, 4, 5])), (deque([2]), deque([1, 4, 5]))]
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