Sorting in bash

Question

I have been trying to get the unique values in each column of a tab delimited file in bash. So, I used the following command.

cut -f <column_number> <filename> | sort | uniq -c

It works fine and I can get the unique values in a column and its count like

105 Linux
55  MacOS
500 Windows

What I want to do is instead of sorting by the column value names (which in this example are OS names) I want to sort them by count and possibly have the count in the second column in this output format. So It will have to look like:

Windows 500
MacOS   105
Linux   55

How do I do this?

Answer 1

Mine:

cut -f <column_number> <filename> | sort | uniq -c | awk '{ print $2" "$1}' | sort

This will alter the column order (awk) and then just sort the output.

Hope this will help you

Answer 2

Use:

cut -f <col_num> <filename>
    | sort 
    | uniq -c
    | sort -r -k1 -n
    | awk '{print $2" "$1}'

The sort -r -k1 -n sorts in reverse order, using the first field as a numeric value. The awk simply reverses the order of the columns. You can test the added pipeline commands thus (with nicer formatting):

pax> echo '105 Linux
55  MacOS
500 Windows' | sort -r -k1 -n | awk '{printf "%-10s %5d\n",$2,$1}'
Windows      500
Linux        105
MacOS         55