how to paste part of the file name to the content of file?

Question

I have a folder with 1000 files; all characters before mapped.ILLUMINA.bwa.GIH.low_coverage.20130415.bam_dp refer to individual's ID (for example NA21117,NA21119,NA21126,..)

NA21117.mapped.ILLUMINA.bwa.GIH.low_coverage.20130415.bam_dp
NA21119.mapped.ILLUMINA.bwa.GIH.low_coverage.20130415.bam_dp
NA21126.mapped.ILLUMINA.bwa.GIH.low_coverage.20121211.bam_dp
NA21127.mapped.ILLUMINA.bwa.GIH.low_coverage.20130415.bam_dp
NA21137.mapped.ILLUMINA.bwa.GIH.low_coverage.20120522.bam_dp
NA21142.mapped.ILLUMINA.bwa.GIH.low_coverage.20130415.bam_dp
NA21143.mapped.ILLUMINA.bwa.GIH.low_coverage.20130415.bam_dp

Each file has only one row:

cat NA21143.mapped.ILLUMINA.bwa.GIH.low_coverage.20130415.bam_dp
1   115258827   10

for each of these files, I want to paste the individual ID to the content of file and get an out put like:

1   115258827   10 NA21143

Is there anyway to do it?

Answer 1

plain bash

for file in *.bam_dp; do 
    contents=$(< "$file")
    echo "$contents ${file%%.*}" > "$file"
done

---

for multi-line files, still can be accomplished with plain bash

for file in *.bam_dp; do 
    mapfile -t contents < "$file"
    printf "%s\n" "${contents[@]/%/ ${file%%.*}}" > "$file"
done

notes:

• the mapfile command reads the file into an array of lines.
• the ${var/pattern/string} parameter expansion does a search-and-replace on the variable value. (documented in the manual)
  • if pattern starts with % the pattern is anchored at the end of the string. Here, I'm matching the empty pattern at the end of the string.
  • the variable can be an array expansions, in which case the replacement occurs for each array element.

Frankly, this approach is too clever, and I'd go for something more obvious.

Answer 2

Use a loop:

#!/bin/bash

shopt -s nullglob
for file in ???????.mapped.*bam_dp; do
  [[ -f "$file" ]] || continue
  id=${file%%.*}              # grab the ID from file name
  sed -i "s/$/ $id/" "$file"  # modify the file in-place
done

Answer 3

Remove .* from $ARGV then append \t $ARGV to the file:

perl -i -pe '$ARGV=~s/\..*//; s/$/\t$ARGV/;' NA*

Glenn's solution is most likely faster to run:

perl -i -lpe '$_ .= " " . substr($ARGV,0,index($ARGV,"."))' NA*

though if each file is only a single line, most of the time will be seeking on the drive.

Answer 4

awk

This method is compatible with the GNU (Linux) and BSD (Mac) versions of awk.

awk '{ id=FILENAME ; sub(/\..*/,"",id) ; print $0 "\t" id }' *.bam_dp

• id=FILENAME ; sub(/\..*/,"",id)
  Store the first part of each *.bam_dp filename (everything before the first .) as id.
• print $0 "\t" id
  Print each file's contents, then a tab character, then the record's id.

This will print a list with lines as in your example:

1   115258827   10 NA21143

The original files will not be modified. You can save this output by, for example, adding > file.txt to the end of the command.