An efficient way to transpose a file in Bash

Question

I have a huge tab-separated file formatted like this

X column1 column2 column3
row1 0 1 2
row2 3 4 5
row3 6 7 8
row4 9 10 11

I would like to transpose it in an efficient way using only bash commands (I could write a ten or so lines Perl script to do that, but it should be slower to execute than the native bash functions). So the output should look like

X row1 row2 row3 row4
column1 0 3 6 9
column2 1 4 7 10
column3 2 5 8 11

I thought of a solution like this

cols=`head -n 1 input | wc -w`
for (( i=1; i <= $cols; i++))
do cut -f $i input | tr $'\n' $'\t' | sed -e "s/\t$/\n/g" >> output
done

But it's slow and doesn't seem the most efficient solution. I've seen a solution for vi in this post, but it's still over-slow. Any thoughts/suggestions/brilliant ideas? :-)

Answer 1

Assuming all your rows have the same number of fields, this awk program solves the problem:

{for (f=1;f<=NF;f++) col[f] = col[f]":"$f} END {for (f=1;f<=NF;f++) print col[f]}

In words, as you loop over the rows, for every field f grow a ':'-separated string col[f] containing the elements of that field. After you are done with all the rows, print each one of those strings in a separate line. You can then substitute ':' for the separator you want (say, a space) by piping the output through tr ':' ' '.

Example:

$ echo "1 2 3\n4 5 6"
1 2 3
4 5 6

$ echo "1 2 3\n4 5 6" | awk '{for (f=1;f<=NF;f++) col[f] = col[f]":"$f} END {for (f=1;f<=NF;f++) print col[f]}' | tr ':' ' '
 1 4
 2 5
 3 6