CODEWITHSUNDEEP
unix
conradlee
conradlee

Reputation: 13695

In UNIX file, quickly sort each line's fields (columns)

I have a file that looks like this

12;6
2;4
9;4
...

In this case the field (column) delimiter is a ";". I want to sort the fields in each line. An acceptable output would be:

6;12
2;4
4;9

An acceptable solution can assume that the field delimiter is a ";" and the values are integers. An ideal solution is more flexible, allowing different delimiters and for alphanumeric sorting.

This all needs to be done on the command line.

Upvotes: 0

Views: 633

Answers (2)

William Pursell
William Pursell

Reputation: 212238

perl -wne '$,=";"; chop; 
    print sort { $a <=> $b } split ";";
    print "\n"' input

If your perl isn't ancient:

perl -wnE '$,=";"; chop; 
        say sort { $a <=> $b } split ";"' input

You can also do:

perl -F\; -wanE 'chop $F[-1]; $,=";"; say sort { $a <=> $b } @F' input

Upvotes: 3

Kent
Kent

Reputation: 195039

 awk 'BEGIN{FS=OFS=";"}{if($1>$2)print $2,$1;else print $1,$2;}' file

test

kent$  cat t.txt
12;6
2;4
9;4
ccc;aaa
bab;baa

kent$  awk 'BEGIN{FS=OFS=";"}{if($1>$2)print $2,$1;else print $1,$2;}' t.txt
6;12
2;4
4;9
aaa;ccc
baa;bab

Upvotes: 2

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