unix for loop modify the output filename

Question

I am working in a directory with file names ending with fastq.gz. with using a loop like the following, I will be running a tool.

for i inls; do if [[ "$i" == *".gz" ]]; then bwa aln ../hg38.fa $i > $i | sed 's/fastq.gz/sai/g'; fi; done

My question is, I want my output filename to end with .sai instead of fastq.gz with keeping the rest of the filename the same. yet, as it first sees $i after >, it modifies the input file itself. I tried using it like <($i | sed 's/fastq.gz/sai/g') but that does not work either. what is the right way of writing this?

Cristian Ramon-Cortes · Accepted Answer · 2017-08-08 09:02:41Z

2

You can use string replacements to compute the filename and the extension. Moreover, you shouldn't rely on the ls output but loop directly over the expression you are looking for.

 for file in *.gz; do
   name="${file%.*}"
   file_output="${name}.sai"
   bwa aln ../hg38.fa ${file} > ${file_output}
 done

answered Aug 8, 2017 at 9:02

Cristian Ramon-Cortes

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2 Comments

Fırat Uyulur Over a year ago

this is a very good way and solves my issue, thanks. just out of curiosity, do you also know how to give the preference to sed on $i and get the modified $i as in the question?

Cristian Ramon-Cortes Over a year ago

Well I will always do it in several steps since it is a script and you do not need to compess it but you could do: CMD > $(echo $file | sed 's/fastq.gz/sai/g')

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