statistical find -exec

I want to run a command over some percentage of the files that find finds. I'll use puts here instead of the real code.

find . -type f -exec ruby -e "puts '{}' if (rand > 0.2)" \;

I have a lot of files. Is there a way to do this without calling rand or even using Ruby?

I thought about using mod X on the f开发者_如何学JAVAilesize, since the files are more or less random length but I can't figure out a way to tell find to use that in an expression.

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This prints 50% of the files:

   find . -exec bash -c "echo -n \$RANDOM | tail -c 1 | grep -q [0-4]" \; -print

To execute something, this should work:

   find . -exec bash -c "echo -n \$RANDOM | tail -c 1 | grep -q [0-4]" \; -exec my_command \;

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I think I found a better way:

find . -type f | ruby -n -e 'puts $_.chop if (rand > 0.5)' 

If you have a lot of files, execing a bash shell for each one can chew up resources. The above line starts one long running ruby process that simply processes the input from find. It's much faster.