Reputation: 13
I would appreciate some help with an awk script, or whatever would do the job.
So, I've got multiple files (the same amount of lines and columns) and I want to do an average of every number in every column (except the first) from all the files. I have got no idea how many columns there are in a file (though i could probably get the number if needed).
filename.1
1 1 2 3 4
2 3 4 5 6
3 2 3 5 6
filename.2
1 3 4 6 6
2 5 6 7 8
3 4 5 7 8
output
1 2 3 5 5
2 4 5 6 7
3 3 4 6 7
I've found this somewhere on here that does it for a single column (as far as I understand it
awk '{a[FNR]+=$2;b[FNR]++;}END{for(i=1;i<=FNR;i++)print i,a[i]/b[i];}' fort.*
So the only? change would be to replace the +=$2 with a cycle over all columns? Is there a way to do that without knowing the exact number of columns?
Thanks.
Upvotes: 1
Views: 632
Reputation: 204731
$ cat tst.awk
{
key[FNR] = $1
for (colNr=2; colNr<=NF; colNr++) {
sum[FNR,colNr] += $colNr
}
}
END {
for (rowNr=1; rowNr<=FNR; rowNr++) {
printf "%s%s", key[rowNr], OFS
for (colNr=2; colNr<=NF; colNr++) {
printf "%s%s", int(sum[rowNr,colNr]/ARGIND+0.5), (colNr<NF ? OFS : ORS)
}
}
}
$ awk -f tst.awk file1 file2
1 2 3 5 5
2 4 5 6 7
3 3 4 6 7
The above uses GNU awk for ARGIND, with other awks just add a line FNR==1{ARGIND++}
at the start.
Upvotes: 4