list unique file extensions recursively for a path, include extension frequency stats

find /some/path -type f -printf '%f\n' | grep -o '\..\+$' | sort | uniq -c | sort -rn
Get the longest match of file extension (Ex. For 'foo.tar.gz', you get '.tar.gz' instead of '.gz')
Sample Output
100 .txt
 25 .tar.gz
 10 .vim
  3 .gitignore

By: skkzsh
2013-03-18 14:42:29

11 Alternatives + Submit Alt

What Others Think

While not strictly identical, because directories with a dot in their name are counted, too, I'd prefer to replace the lengthy find with an ls command: ls -R1A /some/path | grep -o '\..\+$' | sort | uniq -c | sort -rn But this looks still ugly... sort | uniq -c | sort -rn Isn't there a nicer way?
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What do you think?

Any thoughts on this command? Does it work on your machine? Can you do the same thing with only 14 characters?

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