for loop through multiple directories and multiple files

Question

I want to check if each .bam file is accompanied with a .bai file. So if clean_xyz_1.sorted.bam is present clean_xyx_1.sorted.bam.bai should also be present. Each file has a variable string in the middle (xyz). I want to check multiple folders to make sure both files are present. If both files are not present, I want to run a command. However, I have not been able to check for both files in multiple directories. Here is what I have tried:

dirs=(*/)
clean="clean_"
sorted="_1.sorted.bam"

for i in "$dirs"/"$clean"*"$sorted"*; do
  if [[ ! -e "$i".bai ]]; then
  samtools index "$i"
  fi
done

The command works fine and creates a '.bai' file. However, it only opens the first directory. Is there a way to expand all directories?

Answer 1

dir=(*/)

Creates an array of the directory in the current working directory, to iterate an array you should use ${dir[@]} instead of $dir which will only print the first element.

"$dirs"/"$clean"*"$sorted"*

Would also match .bai files. This is probably unwanted behaviour. Hence I suggest using */"$clean"*"$sorted" as the glob of the for loop.

---

Thus I propose this change

shopt -s nullglob
clean="clean_"        
sorted="_1.sorted.bam"
                                          
for i in */"$clean"*"$sorted"; do 
  if [[ ! -e "$i".bai ]]; then
    samtools index "$i"
  fi
done

Answer 2

With zsh, you'd do:

dirs=( *(N/) )
prefix=clean_
suffix=_1.sorted.bam

for file ( $^dirs/$prefix*$suffix(N) )
  [[ -e $file.bai ]] || samtools index $file

Answer 3

Search through dir1, dir2, ... for all .bam files, printing the names of .bai files that should exist but are missing:

find dir1 dir2 ... -type f -name '*.bam' -print | \
while read name ; do
  bai=${name%.bam}.bai
  [ -f "$bai" ] || printf "missing %s\n" "$bai"
done

This assumes you don't have paths with newlines in them, so find outputs one complete .bam pathname per line.