GNU parallel: output each job to a different file without pipes

Question

This question is very close to this other, but that answer is not valid for me, I think due to my shell script does not work with pipes.

This is my multi-job command :

parallel "./ClientesActivos-AP-N.sh -t 15" ::: $(seq 0 2)

I would like output to something like:

file0.out
file1.out
file2.out

I don't know where should I put the redirector >.

I have tested with no luck:

parallel ./ClientesActivos-AP-N.sh -t 15 ">" file{}.out ::: $(seq 0 1)
parallel ./ClientesActivos-AP-N.sh -t 15 ::: $(seq 0 1) ">" file{}.out

My script works in this way:

./ClientesActivos-AP-N.sh -t 15 0
./ClientesActivos-AP-N.sh -t 15 1
./ClientesActivos-AP-N.sh -t 15 2

So output would go (for the above manual unparallelized example) to file0.out, file1.out and file2.out.

What is the correct way to redirect each job to a different file?

Further unsuccessful tests:

parallel --files file{}.out "./ClientesActivos-AP-N.sh -t 15" ::: $(seq 0 2)

What do you mean by *"my script works this way"*? What are the 0, 1 and 2 at the end? — Mark Setchell, May 31 '16 at 00:45
Just the main parameter, @MarkSetchell. I would like this number to be the index. Modified original question to reflect. — Sopalajo de Arrierez, May 31 '16 at 00:47
I think you want `parallel ./ClientesActivos-AP-N.sh -t 15 {} ">" file{#}.out ::: 0 1 2` — Mark Setchell, May 31 '16 at 00:48
@MarkSetchell , your suggestion yields `/bin/bash: ./ClientesActives-AP-N.sh: No such file or directory` three times. Same on absolute path case for the `.sh` script. — Sopalajo de Arrierez, May 31 '16 at 00:50
My spanish spelling was rubbish, please try again. Also try using `parallel --dry-run` to see what it will do but without actually doing anything. — Mark Setchell, May 31 '16 at 00:52
No problem, @MarkSetchell. Here it is: `/ClientesActives-AP-N.sh -t 15 > file1.out` (3 lines with `file2.out` and `file3.out`). — Sopalajo de Arrierez, May 31 '16 at 00:56
LOL. I am a complete idiot!. Excuse me, @MarkSetchell . The method seems to work fine, except for the creation of `file1.out`, 2 and 3 instead of `file0.out`, `file1.out` and `file2.out`. Anyway, it will enough with slight modifications on my script. Thanks a lot. You should write it down as an answer, and if possible expand about the meaning of that `{}`: why is it needed? — Sopalajo de Arrierez, May 31 '16 at 01:11
I found it: just removing the `#` solves the file number issue, @MarkSetchell. — Sopalajo de Arrierez, May 31 '16 at 01:15

score 7 · Accepted Answer · answered May 31 '16 at 07:31

I find that the --dry-run option is a great way to debug GNU Parallel commands. Basically, it tells you what it would do without actually doing anything - it also saves me having to write a dummy "ClientesActivos" script and we all know how good my Spanish isn't ;-)

So, to your immediate question, if you try this, I think what it shows is what you want to do:

$ parallel --dry-run ./ClientesActivos-AP-N.sh -t 15 {} ">" file{}.out ::: {0..1}

./ClientesActivos-AP-N.sh -t 15 0 > file0.out
./ClientesActivos-AP-N.sh -t 15 1 > file1.out

Also have a look at --results: `parallel --results outdir ./ClientesActivos-AP-N.sh -t 15 {} ::: {0..1}` It might be a workable solution for the poster, too. But it does not do exactly what the poster asks for. — Ole Tange, May 31 '16 at 21:50

GNU parallel: output each job to a different file without pipes

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