弹出索引超出范围
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?
解决方案
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
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