Output all column numbers for a particular character

December 23, 2022

I have a matrix(about 10,000×10,000), and I want to find the column number that contains ‘0’.

Matrix (test.txt) :

1 1 1 1 1 1 1 1 1 1
1 0 1 1 1 0 1 1 1 1
1 1 1 0 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1
3 2 2 3 3 0 3 2 2 2
3 2 1 3 3 0 3 2 2 0
3 2 2 3 3 2 3 2 2 2
1 1 1 1 1 1 1 1 1 1

Output (example) :

2 4 6 10

I am new to LINUX SHELL, and have not found much in similar examples. Any help would be much appreciated!!

I just know how to find the row number using grep -nw '0' test.txt|cut -f1 -d':', Maybe I can transpose the matrix first(like this)? And then use the code above, right? Is there an easier way to do it?

>Solution :

$ awk '
    /(^| )0( |$)/ {
        for ( i=1; i<=NF; i++ ) {
            if ( ($i == 0) && !seen[i]++ ) {
                cols[++numCols] = i
            }
        }
    }
    END {
        for ( c=1; c<=numCols; c++ ) {
            printf "%s%s", cols[c], (c<numCols ? OFS : ORS)
        }
    }
' file
2 6 4 10