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Sort file by length of characters in first column / line

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

I have no idea how to do this. I am using Linux so sed / awk / sort are 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

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

.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 it possible?

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3 answers


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

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


If necessary, cut out the auxiliary field with cut -d ' ' -f 2-

.

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You can also do this with coreutils, albeit rather inefficiently:

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

Or with the parallel GNU, if 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-
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or you can also use sed after that,

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

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