列表索引超出范围 [英] list index out of range

问题描述

#!/usr/bin/python

import os,sys
from os import path

input = open('/home/XXXXXX/ERR001268_1', 'r').read().split('\n')

at = 1
for lines in range(0, len(input)):
    line1 = input[lines]
    line4 = input[lines+3]
    num1 = line1.split(':')[4].split()[0]
    num4 = line4.split(':')[4].split()[0]
    print num1,num4


    at += 1

但是我得到了错误:列表索引超出范围

However I got the error: list index out of range

这是什么问题?

btw，除了"at +=1"之外，还有其他方法可以完成此循环吗?

btw, besides "at +=1", is there any other way to finish this cycle loop? thx

推荐答案

让我们说len(input) == 10. range(0, len(input))迭代[0, 1, 2, 3, 4, 5, 6, 7, 8, 9].当行数> 6并且您尝试访问input[lines+3]时，它显然是IndexError，因为没有index[10]，[11]等.

Lets say len(input) == 10. range(0, len(input)) iterates [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]. And when lines > 6 and you're trying to access input[lines+3], it clearly an IndexError, because there is no index[10], [11] etc .

如果line1.count(":") < 4，line1.split(':')[4]也会引发IndexError.

And line1.split(':')[4] can also raise an IndexError if line1.count(":") < 4.

我不了解最后一个at部分，似乎什么也没做，但是您可以使用break语句轻松中断循环.

I didn't understand the last at part, it seems not doing anything, but you can break the loop easily with break statement.

此外，命名变量input是一个坏主意，因为它与内置的input函数冲突.而且range(0, len(input)) == range(len(input))，因此不需要使用0作为范围的第一个参数.

Also, naming a variable input is a bad idea because it conflicts with builtin input function. And range(0, len(input)) == range(len(input)), so 0 as range's first argument is unnecessary.

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