Bash: Sort within a directory, concatenate contents of file with smallest entry in new file - bash

Question

I have a set of directories RUN1, RUN2, etc.
In each directory, there is a set of files. In each file, there are two numbers. For example (these are saved as .csv, even though there are clearly no commas here):
RUN1
mod_1.csv
2.32e-00
1.2e-01
mod_b.csv
4.53e-00
1.1e-01
RUN2
mod_a.csv
1.23e-01
0.5e-02
mod3.csv
1.67e-00
0.4e-01
etc.
I would like to do this:
For each directory:
For each file in a directory:
Sort files by the first entry
Save contents and path of file with smallest value of first entry.
For example, above, this would result in a new file containing:
2.32e-00 1.2e-01 ./RUN1/mod_1.csv
1.23e-01 0.5e-02 ./RUN2/mod_a.csv
I started out by trying this:
#!/bin/bash
resultfile="best_results.txt"
for d in $(find . -type d -name 'RUN*' | sort);
do
find "$d" -type f -name 'mod*' -exec awk '{print $0, FILENAME}' {} \; >> "$resultfile"
done
But it gives me both values from all files, like this:
2.32e-00 ./RUN1/mod_1.csv
1.2e-01 ./RUN1/mod_1.csv
4.53e-00 ./RUN1/mod_b.csv
1.1e-01 ./RUN1/mod_b.csv
1.23e-01 ./RUN2/mod_a.csv
0.5e-02 ./RUN2/mod_a.csv
1.67e-00 ./RUN2/mod_3.csv
0.4e-01 ./RUN2/mod_3.csv
Then I figured I needed to use head, but this modification:
find "$d" -type f -name 'mod*' -exec awk '{print $0, FILENAME}' {} \; | head -1 >> "$resultfile"
gave me:
find: `awk' terminated by signal 13
I think I need another sort, and probably head, but I can't quite put this together.
EDIT (for clarity):
I want to look through all of the files in a directory, find the file with the smallest first number, and write the values of that file and the file path to a new file. Then, proceed to the next directory and do the same. In my example:
Directory RUN1 contains files mod_1.csv and mod_b.csv. File mod_1.csv has the smallest first value. I want to write its contents and file path on one line:
2.32e-00 1.2e-01 ./RUN1/mod_1.csv
to a file.
Directory RUN2 contains files, mod_a.csv and mod3.csv. File mod_a.csv has the smallest first value. I want to write its contents and file path on one line:
1.23e-01 0.5e-02 ./RUN2/mod_a.csv
So that the new file looks like this:
2.32e-00 1.2e-01 ./RUN1/mod_1.csv
1.23e-01 0.5e-02 ./RUN2/mod_a.csv
I understand that this was not clear in my pre-edit question. Please ask any questions you have! I'm not sure how to make this more clear.

Answer 1

You'll probably want to remove the newlines during find from what it looks like:
resultfile="best_results.txt"
for d in $(find . -type d -name 'RUN*');
do
find "$d" -type f -name 'mod*' -exec awk '{printf "%s ",$0} END {print "", FILENAME}' {} \;
done | sort >> "$resultfile"
The sorting would normally done at the very end (once all the results are returned from stdout), however, it's unclear how exactly you expect it to be sorted. You could probably get rid of the for loop if you really wanted, since using something such as the following should work similarly:
find RUN* -type f -name 'mod*' -exec awk '{printf "%s ",$0} END {print "", FILENAME}' {} \; | sort -k 2,2 >> "$resultfile"
Use the -k option with sort to specify which column(s) to sort by.
Result (using sort -k 2,2):
1.67e-00 0.4e-01 RUN2/mod3.csv
1.23e-01 0.5e-02 RUN2/mod_a.csv
4.53e-00 1.1e-01 RUN1/mod_b.csv
2.32e-00 1.2e-01 RUN1/mod_1.csv