Question

This question already has answers here:

Possible Duplicate:
Removing from a list while iterating over it

我有这一法典:

s = [2,3,4,5]

for i in s:
  s.remove(i)

print(s)

在我执政时,结果是:

[3,5]

这里的逻辑错误是什么?

Answer 1

在修改清单时,你正在对清单进行反复讨论,这正造成你的问题。

列出一份临时名单(list(s)),供您根据您的需要公布和修改原件:

>>> s = [2,3,4,5]
>>>
>>> for i in list(s):
...   s.remove(i)
... 
>>> print(s)
[]

Answer 2

合乎逻辑的错误是在通过同一名单删除清单中的内容。

Python hides some implementation details, of course. Internally, you remove the first item, leaving [3,4,5]. Then you remove the second item, leaving [3,5]. At this point you have removed two items from a list that is only two items long, so you are done.

Answer 3

这是因为主持人。 3. 审议本守则部分:

s = [2,3,4,5]

for i in s:
    print "i",i
    j = s
    print "j", j
    s.remove(i)
    print "removed"
    print "j", j
    print "s", s

print s

产出将是

i 2
j [2, 3, 4, 5]
removed
j [3, 4, 5]
s [3, 4, 5]
i 4
j [3, 4, 5]
removed
j [3, 5]
s [3, 5]
[3, 5]

基本上,当你执行居留权时,内部继承人就跑到[0],然后是第3.1条等。当下一轮 lo行时,有1 3.1起已经是4起,而不是3起,因为其中本身已经做了修改。

友情链接