Paste files from list of paths into single output file - linux

Question

I have a file containing a list of filenames and their paths, as in the example below:
$ cat ./filelist.txt
/trunk/data/9.20.txt
/trunk/data/9.30.txt
/trunk/data/50.3.txt
/trunk/data/55.100.txt
...
All of these files, named as X.Y.txt, contain a list of double values. For example:
$ cat ./9.20.txt
1.23
1.0e-6
...
I'm trying to paste all of these X.Y.txt files into a single file, but I'm not sure about how to do it. Here's what I've been able to do so far:
cat ./filelist.txt | xargs paste output.txt >> output.txt
Any ideas on how to do it properly?

Answer 1

You could simply cat-append each file into your output file, as in:
$ cat <list_of_paths> | xargs -I {} cat {} >> output.txt
In the above command, each line from your input file will be taken by xargs, and will be used to replace {}, so that each actual command being run is:
$ cat <X.Y.txt> >> output.txt

Answer 2

If all you're looking to do is to read each line from filelist.txt and append the contents of the file that the line refers to to a single output file, use this:
while read -r file; do
[[ -f "$file" ]] && cat "$file"
done < "filelist.txt" > "output.txt"
Edit: If you know your input file to only contain lines that are file paths (and optionally empty lines) - and no comments, etc. - #Rubens' xargs-based solution is the simplest.
The advantage of the while loop is that you can pre-process each line from the input file, as demonstrated by the -f test above, which ensures that the input line refers to an existing file.

Answer 3

More complex but without argument length limit
Well, the limit here is the available computer memory.
The file buffer.txt must not exist already.
touch buffer.txt
cat filelist.txt | xargs -iXX bash -c 'paste buffer.txt XX > output.txt; mv output.txt buffer.txt';
mv buffer.txt output.txt
What this does, by line:
Create a buffer.txt file which must be initially empty. (paste does not seem to like non-existent files. There does not seem to be a way to make it treat such files as empty.)
Run paste buffer.txt XX > output.txt; mv output.txt buffer.txt. XX is replaced by each file in the filelist.txt file. You can't just do paste buffer.txt XX > buffer.txt because buffer.txt will be truncated before paste processes it. Hence the mv rigmarole.
Move buffer.txt to output.txt so that you get your output with the file name you wanted. Also makes it safe to rerun the whole process.
The previous version forced xargs to issue exactly one paste per file you want to paste but for even better performance, you can do this:
touch buffer.txt;
cat filelist.txt | xargs bash -c 'paste buffer.txt "$#" > output.txt; mv output.txt buffer.txt' FILLER;
mv buffer.txt output.txt
Note the presence of "$#" in the command that bash executes. So paste gets the list of arguments from the list of arguments given to bash. The FILLER parameter passed to bash is to give it a value for $0. If it were not there, then the first file that xargs gives to bash would be used for $0 and thus paste would skip some files.
This way, xargs can pass hundreds of parameters to paste with each invocation and thus reduce dramatically the number of times paste is invoked.
Simpler but limited way
This method suffer from limitations on the number of arguments that a shell can pass to a command it executes. However, in many cases it is good enough. I can't count the number of times when I was performing spur-of-the-moment operations where using xargs would have been superfluous. (As part of a long term solution, that's another matter.)
The simpler way is:
paste `cat filelist.txt` > output.txt
It seems you were thinking that xargs would execute paste output.txt >> output.txt multiple times but that's not how it works. The redirection applies to the entire cat ./filelist.txt | xargs paste output.txt (as you initially had it). If you want to have redirection apply to the individual commands launched by xargs you have it launch a shell, like I do above.

Answer 4

#!/usr/bin/env bash
set -x
while read -r
do
echo "${REPLY}" >> output.txt
done < filelist.txt
OR, to get the files directly:-
#!/usr/bin/env bash
set -x
find *.txt -type f | while read $files
do
echo "${files}" >> output.txt
done

Answer 5

A simple while loop should do the trick:
while read line; do
cat ${line} >> output.txt
done < filelist.txt