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]

Rather than do extra work to figure out the indices and then apply them back to the other list, why not pair the values up before sorting (for example, using `zip`)? — Karl Knechtel, Sep 13 '22 at 18:48

score 50 · Answer 1 · answered Jul 15 '10 at 22:04

50

[listofLines[i] for i in sortedIndex]

answered Jul 15 '10 at 22:04

sepp2k

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score 11 · Accepted Answer · answered Jul 15 '10 at 22:04

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)

score 3 · Answer 3 · answered Jul 15 '10 at 22:02

3

sorted(zip(listofTimes, listofLines))

answered Jul 15 '10 at 22:02

Anycorn

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K3---rnc · Answer 4 · 2021-09-25T16:05:19.763

1

If you want to sort the original list because you, say, hold references to it elsewhere, you can assign to it the sorted list:

my_list[:] = [my_list[i] for i in sorted_indexes]  # [:] is key!

edited Sep 25 '21 at 16:05

answered Jun 13 '16 at 23:01

K3---rnc

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Sort list by given order of indices

4 Answers4

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