Sort list by given order of indices

Question

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.

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.

Example:

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]

Answer 1

[listofLines[i] for i in sortedIndex]

Answer 2

I think you could do

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

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

def extract_time(line):
    ...
    return time

you could also do

listofLines.sort(key=extract_time)

or if you want to keep the original list intact,

sorted(listofLines, key=extract_time)