Using bash to sort data horizontally

I have a file full of data in columns

sarah mark john
10    20   5
x     y    z

I want to sort the data so the columns stay intact but the second row is in increasing order so it looks like this:

john sarah mark
5    10    20 
z    x     y

I've been looking at the sort command but have only been able to find vertical sorting, not horizontal. I'm happy to use any tool, any help is appreciated. Thank you!

Let's create a function to transpose a file (make rows become columns, and columns become rows):

transpose () {
  awk '{for (i=1; i<=NF; i++) a[i,NR]=$i; max=(max<NF?NF:max)}
        END {for (i=1; i<=max; i++)
              {for (j=1; j<=NR; j++) 
                  printf "%s%s", a[i,j], (j<NR?OFS:ORS)
              }
        }'
}

This just loads all the data into a bidimensional array a[line,column] and then prints it back as a[column,line], so that it transposes the given input. The wrapper transpose () { } is used to store it as a bash function. You just need to copy paste it in your shell (or in ~/.bashrc if you want it to be a permanent function, available any time you open a session).

Then, by using it, we can easily solve the problem by using sort -n -k2: sort numerically based on column 2. Then, transpose back.

$ cat a | transpose | sort -n -k2 | transpose
john sarah mark
5 10 20
z x y

In case you want to have a nice format as final output, just pipe to column like this:

$ cat a | transpose | sort -n -k2 | transpose | column -t
john  sarah  mark
5     10     20
z     x      y

Step by step:

$ cat a | transpose 
sarah 10 x
mark 20 y
john 5 z
$ cat a | transpose | sort -n -k2
john 5 z
sarah 10 x
mark 20 y
$ cat a | transpose | sort -n -k2 | transpose 
john sarah mark
5 10 20
z x y

Coming from a duplicate question, this would sort the columns by the first row:

#!/bin/bash

input="$1"

order=$((for i in $(head -1 $input); do echo $i; done) | nl | sort -k2 | cut -f1)

grep ^ $input | (while read line
  do
    read -a columns <<< "${line%"${line##*[![:space:]]}"}"

    orderedline=()
    for i in ${order[@]}
    do
      orderedline+=("${columns[$i - 1]}")
    done
    line=$(printf "\t%s" "${orderedline[@]}")
    echo ${line:1}
  done)

To sort by second row, replace head -1 $input with head -2 $input | tail -1. If the sort should be numeric, put in sort -n -k2 instead of sort -k2.

Good one-liner gets the job done:

perl -ane '$,=" "; print sort @F; print "\n";' file

I found it here: http://www.unix.com/unix-for-advanced-and-expert-users/36039-horizontal-sorting-lines-file-sed-implementation.html

Using bash to sort data horizontally

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