Bash retrieve column number from column name

Question

Is there a better way (such as a one liner in AWK) where I can get the column number in a table with headings from a column name? I want to be able to process a column independent of what the column number actually is (such as when another column is added the script will not need to change).

For example, given the following table in "table.tsv":

ID  Value   Target  Not Used
1   5   9   11
2   4   8   12
3   6   7   10

I can do a sort on the "Target" column using:

#!/bin/bash
(IFS=$'\t'; read -r; printf "%s\n" "$REPLY"; i=0; for col in $REPLY; do
    ((++i))
    [ "$col" == "Target" ] && break
done; sort -t$'\t' "-k$i,${i}n") < table.tsv

Is there a way to do it without the for loop (or at least clean it up a little)?

The expected output of the given script is:

ID      Value   Target  Not Used
3       6       7       10
2       4       8       12
1       5       9       11

However, I was trying to give an example of what I was trying to do. I want to pass/filter my table through several programs so the headings and all columns should be preserved: just have processing occur at each step. In pseudo code, what I would like to do is:

print headings from stdin
i=$(magic to determine column position given "Target")
sort -t$'\t' "-k$i,${i}n"  # or whatever processing is required on that column

Answer 1

another alternative with a lot of pipes

$ head -1 table | tr -s ' ' '\n' | nl -nln |  grep "Target" | cut -f1

extract first row, transpose, number lines, find column name, extract number

Or, awk to the rescue!

$ awk -v RS='\t' '/Target/{print NR; exit}' file.tsv
3