I want to free up some space by deleting redundant files within many subdirectories. However I always want to retain the last file of a file stem. For example I have an assortment (next to other files that should not be touched) similar to this:

FILE_1-1 FILE_1-2 FILE_1-3 FILE_2-1 FILE_2-2

within a whole bunch of branching subdirectories. I want to keep FILE_2-2 in this instance, but in another folder it would be FILE_3-1 or whatever. Always the last file. I know how to get rid of all of them using the find command which traverses the subdirectories nicely:

find . -name "FILE_*" -type f -delete

How can I exclude the last FILE_* within each directory where removal happens using the find command? Googling (https://www.unix.com/shell-programming-and-scripting/36639-find-remove-all-but-latest-file.html) turned up the following:

ls -1 FILE_* | sort -r | tail +2 | xargs -i rm {}

But this command does not work with branching directories, does it?

  • to make matters worse, you'll also need to worry about the sort functionality if you have more than one digit in the sequence. FILE_1-10 comes lexicographically before FILE_1-2 . Padding with zeros (eg FILE_1-002 ) would make numeric and lexicographic order consistent.
    – erik258
    Sep 27, 2019 at 15:04
  • For a general answer, yes. In my case the highest number is fortunately FILE_3-5. I hope this helps to keep it simple.
    – And
    Sep 27, 2019 at 15:07
  • You mention branching subdirectories, but they need to be considered when finding "the last file", or is the "last file" the one in each separate subdirectory?
    – Jeff Schaller
    Sep 27, 2019 at 16:13
  • There's a slight inconsistency; your example command shows ls ... FILE_*.txt while your sample filenames do not end in .txt
    – Jeff Schaller
    Sep 27, 2019 at 16:16
  • You are correct, I edited that out. And It's always the last file within each separate subdirectory.
    – And
    Sep 30, 2019 at 8:01

1 Answer 1


In a directory you can find the "stems" with:

ls * | sed -r 's/[0-9]+$//' | sort | uniq

For each stem you can list all files with the last with:

ls  FILE_1-* | head -n -1

Or if you want to keep the most recent:

ls -rt FILE_1-* | head -n -1

Both commands can be followed by | xargs rm to delete their output

Put all that in a short script(*) and run:

find . -type d -exec {the_script} {} \;

(*) the short script would look like:

#! /bin/bash
cd "$1"  # takes the directory as a parameter
for stem in $(ls FILE* | sed -r 's/[0-9]+$//' | sort | uniq)
    ls -rt ${stem}* | head -n -1 | xargs echo rm 

Of course test thoroughly and replace echo rm by just rm when you are confident that it will erase the right files.

  • I put the first and the third line into the script and then executed the last line and I'm getting the error find: ‘{the_script}’: No such file or directory. I put the script into the parent folder where I am executing the find command. What do I do wrong?
    – And
    Sep 30, 2019 at 10:24
  • {the script} is of course a place holder for your actual script name. Note that your script is also assumed to loop over the stems: line 3 is executed once for each stem found with line 1.
    – xenoid
    Sep 30, 2019 at 11:17
  • Added sample script, use at your own risk.
    – xenoid
    Sep 30, 2019 at 11:40

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