Sort based on column

sort

10 awhdk;
14 hjoeow;
2  kdkld;
4  jkjdksl

How to sort this based on column one (i.e the count) ?

I'd want to get this output :

14 hjoeow;10 awhdk;4 jkjdksl;2 kdkld

I need both the columns to be printed based on sorting the first column.

Best Answer

You can use -k to specify where in the input line the sort key starts and ends.

You use the -n option or add n to the sort key specification to trigger a numeric sort as opposed to a lexical sort by default. And -r to reverse the sort.

Here, for a numerical sort, you don't need to limit the sort key to the sort command. If you do:

sort -rn file

It will treat the whole line as a number. The 14 hjoeow; line for instance will be understood as the 14 number.

You could do:

sort -bk 1,1rn file

To sort on the first field only, but that would make no difference.

The command to join lines is paste with the -s option. You can specify the delimiter with -d. \0 is a special value for the delimiter that means join with nothing in between. So, to get your expected output:

sort -rn file | paste -sd '\0' -

POSIX paste require a filename be passed. - is again a special value that means standard input. You can omit it with some paste implementations.

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