python:迭代弹出元素时列出索引超出范围错误 [英] python : list index out of range error while iteratively popping elements

问题描述

我已经编写了一个简单的python程序

I have written a simple python program

l=[1,2,3,0,0,1]
for i in range(0,len(l)):
       if l[i]==0:
           l.pop(i)

如果l [i] == 0:

This gives me error 'list index out of range' on line if l[i]==0:

调试后，我发现 i 越来越多，列表也越来越少.
但是，我有循环终止条件 i<len(l).那为什么我会收到这样的错误?

After debugging I could figure out that i is getting incremented and list is getting reduced.
However, I have loop termination condition i < len(l). Then why I am getting such error?

推荐答案

在迭代列表时，您正在减少列表 l 的长度，因此当您在索引中接近索引的末尾时范围语句，其中一些索引不再有效.

You are reducing the length of your list l as you iterate over it, so as you approach the end of your indices in the range statement, some of those indices are no longer valid.

看起来就像您想要做的一样:

It looks like what you want to do is:

l = [x for x in l if x != 0]

将返回 l 的副本，其中没有任何为零的元素(该操作称为

which will return a copy of l without any of the elements that were zero (that operation is called a list comprehension, by the way). You could even shorten that last part to just if x, since non-zero numbers evaluate to True.

不存在 i<的循环终止条件.len(l)，就像您编写代码的方式一样，因为 len(l)是在循环之前 pre 计算的，因此不会重新评估每次迭代.您可以以这种方式编写它，但是:

There is no such thing as a loop termination condition of i < len(l), in the way you've written the code, because len(l) is precalculated before the loop, not re-evaluated on each iteration. You could write it in such a way, however:

i = 0
while i < len(l):
   if l[i] == 0:
       l.pop(i)
   else:
       i += 1

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