CODEWITHSUNDEEP
bashshell
elcortegano
elcortegano

Reputation: 2684

Naming an output file after the input directory

I'm working with some files that are organized within a folder (named RAW) that contain several other folders with different names, all of them containing files ended by a string like _1 or _2 with the extension (.fq.gz in this case). Below I try to include a schedule for guidance.

RAW/
  FOLDER1/
    FILE_qwer_1.fa.gz
    FILE_qwer_2.fa.gz
  FOLDER2/
    FILE_tyui_1.fa.gz
    FILE_tyui_2.fa.gz
  OTHER1/
    FILE_asdf_1.fa.gz
    FILE_asdf_2.fa.gz
  ...

So I am basically running a loop over all those directories under RAW and run a script that will create an output file, say out.

What I'm trying to accomplish is to name that out file as the folder it belongs to under $RAW (e.g. FOLDER1.eg after processing FILE_qwer_1.fa.gz and FILE_qwer_2.fa.gz above)

The loop below will work actually, but as you can imagine, it depends on how many folders I am working below the root /, as the option -f is hard-coded for the cut command.

for file1 in ${RAW}/*/*_1.fq.gz; do
        file2="${file1/_1/_2}"
        out="$(echo $file1 | cut -d '/' -f2)"
        bash script_to_be_run.sh $file1 $file2 $out
done

Ideally, the variable out should be named as the replacement of the first * character of the glob used in the loop (e.g. FOLDER1.eg in the first iteration) followed by a custom extension, but I do not really know how to do it, nor if it is possible.

Upvotes: 1

Views: 97

Answers (1)

ntki
ntki

Reputation: 2304

You can use ${var#prefix} to remove a prefix from the start of a variable.

for file1 in ${RAW}/*/*_1.fq.gz; do
  file2="${file1/_1/_2}"
  out="$(dirname "${file1#$RAW/}")"  # cuts the $RAW from the beginning of the dirs
  bash script_to_be_run.sh "$file1" "$file2" "$out"
done

(It's a good idea to quote variable expansions in case they contain spaces or other special character: "$file1" is safer than $file1.)

Upvotes: 2

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