问题描述

N=8
f,g=4,7
indexList = range(N)
print indexList
print f, g
indexList.pop(f)
indexList.pop(g)

在此代码中，我收到一条错误消息，指出 indexList 中 g 的弹出索引超出范围.这是输出:

In this code I am getting an error stating that the pop index of g in indexList is out of range. Here is the output:

[0, 1, 2, 3, 4, 5, 6, 7]
4 7
Traceback (most recent call last):
indexList.pop(g)
IndexError: pop index out of range

我不明白，g 的值为 7，列表包含 7 个值，为什么无法返回列表中的 7?

I don't understand, g has a value of 7, the list contains 7 values, why is it not able to return me the 7 in the list?

推荐答案

要获取弹出列表的最终值，可以这样:

To get the final value of a list pop'ed, you can do it this way:

>>> l=range(8)
>>> l
[0, 1, 2, 3, 4, 5, 6, 7]
>>> l.pop(4)                    # item at index 4
4
>>> l
[0, 1, 2, 3, 5, 6, 7]
>>> l.pop(-1)                   # item at end - equivalent to pop()
7
>>> l
[0, 1, 2, 3, 5, 6]
>>> l.pop(-2)                   # one left of the end 
5
>>> l
[0, 1, 2, 3, 6]
>>> l.pop()                     # always the end item
6
>>> l
[0, 1, 2, 3]

请记住，pop 会删除该项目，并且列表在弹出后更改长度.使用负数从可能改变大小的列表末尾开始索引，或者只使用不带参数的 pop() 结束项.

Keep in mind that pop removes the item, and the list changes length after the pop. Use negative numbers to index from the end of a list that may be changing in size, or just use pop() with no arguments for the end item.

由于 pop 会产生这些错误，因此您经常会在 异常块:

Since a pop can produce these errors, you often see them in an exception block:

>>> l=[]
>>> try:
...    i=l.pop(5)
... except IndexError:
...    print "sorry -- can't pop that"
... 
sorry -- can't pop that

这篇关于弹出索引超出范围的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持跟版网！