Naming output files based on variable parameters and input filenames

Shell Programming and Scripting
1

Hello, I have a series of files in sub-directories that I want to loop through, process and name according to the input filename and the various parameters I'm using to process the files. I have a number of each, for example file names like AG005574, AG004788, AG003854 and parameter values like ATd, PZa, RTK1, so I want to end with files like AG005574_ATd, AG005574_PZa, AG005574_RTK1, AG004788_ATd, etc.
I loop through the subfolders, run the process and output the results like so:

#!/usr/bin/bash
model=1
output=$file
for file in $(find /path/to/files/AG00*/*/ -type f -name 'AG00*.fa');
 do process --out=$output."tab" <options> $file ;
      echo $file
done

Obviously, this names the file by process only, not by process_input, and it only writes one file (the last processed) because it over-writes the others. In the case above, the --out= seems to me to be the key to change to get what I want, but $output$file and variations don't work.
I'm sure this is not a tricky problem but I haven't been able to find or piece together a solution. Any help/tutoring is much appreciated.

2

output=$file is empty.
That is because it is outside the loop, try to place it 2 lines lower.
Therefor both, the current and your 'future' line:

 do process --out=$output."tab" <options> $file ;

Looks in plaintext:

 do process --out=."tab" <options> Value_from_Find ;

Or at best (with the thought change):

 do process --out=Value_from_Find."tab" <options> Value_from_Find ;

For the moment, try:

#!/usr/bin/bash
model=1
for file in $(find /path/to/files/AG00*/*/ -type f -name 'AG00*.fa');
 do output=$file
      process --out=$output."tab" <options> $file ;
      echo $file
done

Hope this helps

EDIT:
model=1 is not used and could be deleted.

3

Thanks, Sea.

I should have been more explicit about why I used

model=1

in the <options> for the process I run, I point to several different models for processing, like so:

process --out=$output."tab" /path/to/input/models/$1.hmm $file;

so that when I have

--out=$model."tab" I get part of the filename I want for my output (namely, the model I used) but not the name of the input file that was processed, and each new file that is processed over-writes the previous.

when I try something like

--out=$model$output."tab"

or

--out=$model_$output."tab"

or

--out=$model_$file."tab"

the process/program errors. The solution you suggested also causes an error message from the program, "no such option --tblfp", which makes no sense to me.