Bash: sort text file by last field value

Question

Hi All,

I have a text file containing ~300k rows. Each row has a varying number of comma-delimited fields, the last of which is guaranteed numerical. I want to sort the file by this last numerical field. I can't do:

sort -t, -n -k 2 file.in > file.out

as the number of fields in each row is not constant. I think sed, awk maybe the answer, but not sure how. E.g:

awk -F, '{print $NF}' file.in

gives me the last column value, but how to use this to sort the file?

Answer 1

Maybe reverse the fields of each line in the file before sorting? Something like

perl -ne 'chomp; print(join(",",reverse(split(","))),"\n")' |
  sort -t, -n -k1 |
  perl -ne 'chomp; print(join(",",reverse(split(","))),"\n")'

should do it, as long as commas are never quoted in any way. If this is a full-fledged CSV file (in which commas can be quoted with backslash or space) then you need a real CSV parser.

Answer 2

vim file.in -c '%sort n /.*,\zs/' -c 'saveas file.out' -c 'q'

Answer 3

Use awk to put the numeric key up front. $NF is the last field of the current record. Sort. Use sed to remove the duplicate key.

awk -F, '{ print $NF, $0 }' < yourfile | sort -n -k1 | sed 's/^[0-9][0-9]* //'

Answer 4

Perl one-liner:

@lines=<STDIN>;foreach(sort{($a=~/.*,(\d+)/)[0]<=>($b=~/.*,(\d+)/)[0]}@lines){print;}