按给定的索引顺序对列表进行排序 [英] Sort list by given order of indices

问题描述

我有一个从文件中读取的行的列表.我需要按时间戳对列表进行排序.我已经使用正则表达式解析了时间戳，并将其放置在单独的列表中.这两个列表的索引将匹配.对时间戳列表进行排序后，便可以获取索引的顺序.

I have a list of lines read from a file. I need to sort the list by time stamp. I have parsed out the time stamp using regular expressions and place them into a separate list. The indices of the two lists will match. Once I sort the list of time stamps, I can get the order of indices.

是否可以将相同顺序的索引应用于原始行列表?结果应该是原始行的排序列表.

Is there a way to apply the same order of indices to the original list of lines? The result should be the sorted list of original lines.

示例:

listofLines =  ['log opened 16-Feb-2010 06:37:56 UTC', 
                '06:37:58 Custom parameters are in use',
                'log closed 16-Feb-2010 05:26:47 UTC']
listofTimes = ['06:37:56', '06:37:58', '05:26:47']
sortedIndex = [2,0,1]

推荐答案

我认为您可以做到

[line for (time,line) in sorted(zip(listofTimes, listofLines))]

但是，如果您具有(或可以编写)自动从行中提取时间的功能，

But if you have (or could write) a function to automatically extract the time from the line,

def extract_time(line):
    ...
    return time

您也可以

listofLines.sort(key=extract_time)

或者如果您想保持原始列表不变，

or if you want to keep the original list intact,

sorted(listofLines, key=extract_time)

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