Bash: replace part of filename

Question

I have a command I want to run on all of the files of a folder, and the command\'s syntax looks like this:

tophat -o  


        

Answer 1

Alternative to anubhava's concise solution,

d=$(dirname path/to/sample1.fastq)
b=$(basename path/to/sample1.fastq .fastq)
echo $d/$b.fastq
path/to/sample1.fastq

tophat -o "$d/$b.fastq" "$f"

Answer 2

Not an answer but a suggestion: as a bioinformatician, you shoud use GNU make and its option -j (number of parallel jobs). The Makefile would be:

.PHONY:all
FASTQS=$(shell ls *.fastq)

%.bam: %.fastq
    tophat -o $@ $<

all:  $(FASTQS:.bam=.fastq)

Answer 3

You can use:

tophat -o "${f/.fastq/.bam}" "$f"

Testing:

f='path/to/sample1.fastq'
echo "${f/.fastq/.bam}"
path/to/sample1.bam