pair up files with matching extensions

Let's say I have this list of files:

a.1

a.2

a.3

开发者_C百科 b.1

b.2

Is there a bash one-liner that could find the one 'a' file for which there is no 'b' with the same extension? (i.e. a.3: no match)

I'm sure I could write a short bash/perl script to do this.

But I would like to know if there is any "trick" for this (of course, GNU tools are at my disposal; awk, sed, find ...)

You could try this:

ls -1 [ab].* | sort -t . -k 2 | uniq -u -s2

Here is my bash one-liner

for afile in a*; do bfile=${afile/a/b}; test -f $bfile || echo $afile; done

I like the above solution since it only uses bash and no other tools. However, to showcase the power of Unix tools, the next solution uses 4: ls, sed, sort, and uniq:

ls [ab]* | sed 's/b/a/' | sort | uniq -u

If you can use Perl:

perl -le 'for (<a*>) { /(.*)\.(.*)/; print "$1.$2" if !-e "b.$2"}'

bash version 4 has associative arrays, so you can do this:

declare -A a_files
while read -r filename; do
    ext="${filename##*.}"
    case "${filename%.*}" in 
        a) a_files[$ext]="$filename" ;;
        b) unset a_files[$ext] ;;
    esac
done < <(ls [ab].*)
echo "non-matched 'a' files: ${a_files[@]}"

Or, with awk:

ls [ab].* | awk -F. '
    $1 == "a" {a_files[$2] = $0} 
    $1 == "b" {delete a_files[$2]}
    END {for (ext in a_files) print a_files[ext]}
'

Ruby(1.9+)

$ ruby -e 'Dir["a.*"].each{|x|puts x if not File.exist? "b."+x.scan(/a\.(\d+)/)[0][0]}'