sort a file based on length of chars in the first column/row

4
votes

I need to sort a file based on the number of chars in the first column.

I have no idea on how to go about this. I'm using Linux, so sed/awk/sort are all available.

.abs is bla bla 12 
.abc is bla se 23 bla
.fe is bla bla bla
.jpg is pic extension
.se is for swedish domains

What I want is to sort these lines, based on the length of the first column in each line. Some of the lines start with 4 characters, some start with 3, or 2. I want the result to be something like:

.fe is bla bla bla 
.se is for swedish domains 
.abs is bla bla 12 
.abc is bla se 23 bla 
.jpg is pic extension

Is this even possible?

3
linuxsortingsedawkgrep
for the lines with same length (column1), how would you sort them? - Kent

3 Answers

10
votes

Augment each line by the length of the first word, then sort:

awk '{ print length($1) " " $0; }' $FILE | sort -n

If necessary, cut out the helper field with cut -d ' ' -f 2- afterwards.

2
votes

or you can also use sed afterwards like this

awk '{print length($1)" "$0}' temp.txt | sort -k 1,2| sed -re 's/^[0-9]+ //'

1
votes

You can also do it with coreutils, albeit rather inefficient:

paste -d' ' <(cut -d' ' -f1 infile | xargs -l sh -c 'echo "$1" | wc -c' '{}') infile |
  sort -n | cut -d' ' -f2-

Or with GNU parallel if it is available:

paste -d' ' <(cut -d' ' -f1 infile | parallel wc -c '<<< {}') infile | 
  sort -n | cut -d' ' -f2-

Or with bash:

<infile while read c1 rest; do echo ${#c1} "$c1" "$rest"; done |
  sort -n | cut -d' ' -f2-