Executing command on every file in folder in random order

Question

I want to execute one command for every file in a folder, but randomly. Something like:

find . -type f -exec <command> '{}' \;

but every time in another order. This is closest to my needs, but 1) it doesn't work, 2) order is random, but always the same:

find . -type f -print0 | sort -Rz | xargs -0 <command>

Answer 1

find . -type f -exec <command> '{}' \;

is not equivalent to

find . -type f | xargs <command>

observe:

$ find -type f
./b
./c
./e
./d
./a
$ find -type f -exec echo '{}' \;
./b
./c
./e
./d
./a
$ find -type f | xargs echo
./b ./c ./e ./d ./a

xargs collects a bunch of arguments up to a given length and then executes the command with all arguments at once.

This is what gets executed from find -type f -exec echo '{}' \;

echo ./b
echo ./c
echo ./e
echo ./d
echo ./a

This is what gets executed from find -type f | xargs echo

echo ./b ./c ./e ./d ./a

This works well for commands that can take multiple arguments like md5sum or file. But does not work for commands that only take one argument at a time.

To make xargs behave more like find -exec you can add argument -n1 to xargs:

$ find -type f | xargs -n1 echo
./b
./c
./e
./d
./a

-n1 tells xargs to execute a command for every 1 arguments.

In your example command:

find . -type f -print0 | sort -Rz | xargs -0 -n1 <command>

---

Bonus: you can also make find -exec to behave more like xargs by closing the -exec with \+ instead of \;:

find -type f -exec <command> '{}' \+