ls *.txt | parallel 'echo Starting on file {}; mkdir {.}; cd {.}; longCMD3 ../{} > /dev/null; echo Finished file {}'

This one liner partially works, except longCMD3 takes about 3 minutes, but the first and second echo commands are printed almost at the same time. I tried putting in


before the final echo, but that made no difference.

How can I ensure that the final echo is only printed once longCMD3 is complete?

Here's an example

Assume I only have 4 cores:

foo1.txt foo2.txt foo3.txt foo4.txt foo5.txt foo6.txt 

What I expected:

Starting on file foo1.txt
Starting on file foo2.txt
Starting on file foo3.txt
Starting on file foo4.txt

then at least 2 minutes should pass for longCMD3 to finish on one of the files

Finished file foo1.txt
Starting on file foo5.txt

But what I get is:

Starting on file foo1.txt
Finished file foo1.txt
Starting on file foo2.txt
Finished file foo2.txt
Starting on file foo3.txt
Finished file foo3.txt
Starting on file foo4.txt
Finished file foo4.txt

This continues for all 6 files. And the Start and Finished statements are printed simultaneously for each file. But a few minutes are expended between each file.

  • 4
    It isn't clear what you want here. For each .txt file, echo Finished file foo.txt runs after longCMD3 ../foo.txt has finished. But it's to be expected that “Finished” for one file appears before all the files have been processed: otherwise why would you parallelize? Dec 29, 2017 at 14:41
  • OK, I hope it's clear now.
    – Glubbdrubb
    Dec 29, 2017 at 15:39
  • If the finishes are arriving approximately on time, but the starts are arriving just before the finishes, and the process monitor is saying that you have 8 processes running for sometime (Please confirm). Then it is probably an output buffer problem. That is that the output is being buffered, and output all at the same time. This may be by design, to so that outputs are not mixed up. Dec 29, 2017 at 16:00
  • @ctr-alt-del, you are indeed correct. As mentioned by Gilles, the output was being buffered by parallel.
    – Glubbdrubb
    Dec 29, 2017 at 16:23

1 Answer 1


For each file, the commands echo Starting on file foo.txt, mkdir foo, cd foo, longCMD3 ../foo.txt > /dev/null and echo Finished file foo.txt run sequentially, i.e. each command starts after the previous one has finished.

The commands for different files are interspersed. By default, the parallel command runs as many jobs in parallel as you have cores.

However the output of the commands is not interspersed by default. This is why you don't see a bunch of “Starting” lines and then later the corresponding “Finished” lines. Parallel groups the output of each job together. It buffers the output until the job is finished. See the description of the --group option in the manual. Grouping doesn't make sense in your case, so turn it off with the --ungroup (-u) option, or switch to line grouping with --line-buffer.

Some other corrections:

  • Parsing ls is not reliable. Pass the file names to parallel directly.
  • If mkdir fails, you shouldn't proceed. If any command fails, you should arrange for the job to fail. An easy way to do that is to start the job script with set -e.
parallel --line-buffer 'set -e; echo Starting on file {}; mkdir {.}; cd {.}; longCMD3 ../{} > /dev/null; echo Finished file {}' ::: *.txt
  • The -u option worked beautifully. And I'll also change the way the file names are passed to parallel.
    – Glubbdrubb
    Dec 29, 2017 at 16:28
  • @Glubbdrubb -u does not guarantee half-lines to mix. It is unlikely to happen in your case (because your lines are < 4 KB long), but I suppose --line-buffer is really what you are looking for.
    – Ole Tange
    Dec 30, 2017 at 14:18

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