Bash: sorting by two fields

humbletech99 · 09-24-2007, 08:24 AM

I am using bash and need to sort a bunch of numbers that look like the following example:

Code:

9.1
9.2
...
9.N
10.1
10.2
...
10.N

I am having trouble because the sort command either sorts by one field or the other, so if I just | sort -n I end up with

Code:

9.1
9.20
9.21
...
9.2
9.30
9.31
...
9.3
...
10.1
10.20
10.21
...
10.2

Does anyone have a cunning way of sorting by field one and then subsorting by field 2 without losing the order of the first field? If I sort by field 2 then of course the 9 and 10 at the beginning would could out of order.

druuna · 09-24-2007, 08:36 AM

Hi,

Did you try the -g option instead of the -n option?

Man sort for the details.

Hope this helps.

humbletech99 · 09-24-2007, 09:02 AM

yes of course I manned sort before posting and I tried -g as well at that time but it doesn't give the desired effect, the sort still comes out wrong with .11 before .2 for example, same as demonstrated above.

druuna · 09-24-2007, 09:15 AM

Unusefull (read: dumb) reply removed.

matthewg42 · 09-24-2007, 09:24 AM

Well, you can do it with a short Perl program. Be aware that this will slurp up the whole input file into memory, so if it's a huge file, this is probably not a good method.

Code:

#!/usr/bin/perl
# save this into a file called "mysort" and make that executable
# then invoke "mysort" on your input file.  The sorted output will
# be printed on stdout, so you can re-direct it like this:
#
# mysort  input_file > output_file

sub my_cmp {
        @a=split(/\./, $a, 3);
        @b=split(/\./, $b, 3);

        if ($a[0] ne $b[0]) { return $a[0] <=> $b[0]; }
        else { return $a[1] <=> $b[1]; }
}

foreach (sort my_cmp <>) { print; }

Hobbletoe · 09-24-2007, 12:50 PM

For just bash, you can try ...

Code:

#!/bin/bash

list=($(cat infile | awk -F . '{print $1}' | sort -ug))

for i in $(seq 0 $((${#list[*]}-1)))
do 
 ftemp=$(mktemp)
 egrep ^${list[${i}]}'\.' infile | awk -F . '{print $2}' >> ${ftemp}
 list2=($(sort -n ${ftemp}))
 for j in $(seq 0 $((${list2[*]}-1)))
 do
  echo ${list[${i}]}.${list2[${j}]}
 done
 rm ${ftemp}
done

Basically, what I've done is to sort the first field (list), then create a temporary file that contains only the second field for each list, and I sort that out into a second list (list2). Then it is a simple matter of echoing out the first field . second field. Then get rid of the temporary file at the end of each first field, though I suppose you could skip that if you want. This works if you have something like 1 and 10 in the same file as well (set my test data up to make sure).

Hobbletoe Clubfoot

LasseW · 09-25-2007, 01:29 PM

Assuming the numbers are in file.dat:

$ awk -F. '{print $1,$2}' file.dat|sort -nk1 -nk2|sed 's/ /./'
9.1
9.2
9.3
9.11
9.20
9.21
9.30
9.31
10.1
10.2
10.11
10.20
10.21

makyo · 09-25-2007, 03:30 PM

Hi.

See http://forums.debian.net/viewtopic.php?t=19478 ... cheers, makyo