How to sort files in paste command with 500 files csv - shell

Question

My question is similar to How to sort files in paste command?
- which has been solved.
I have 500 csv files (daily rainfall data) in a folder with naming convention chirps_yyyymmdd.csv. Each file has only 1 column (rainfall value) with 100,000 rows, and no header. I want to merge all the csv files into a single csv in chronological order.
When I tried this script ls -v file_*.csv | xargs paste -d, with only 100 csv files, it worked. But when tried using 500 csv files, I got this error: paste: chirps_19890911.csv: Too many open files
How to handle above error?
For fast solution, I can divide the csv's into two folder and do the process using above script. But, the problem I have 100 folders and it has 500 csv in each folder.
Thanks
Sample data and expected result: https://www.dropbox.com/s/ndofxuunc1sm292/data.zip?dl=0

Answer 1

You can do it with gawk like this...
Simply read all the files in, one after the other and save them into an array. The array is indexed by two numbers, firstly the line number in the current file (FNR) and secondly the column, which I increment each time we encounter a new file in the BEGINFILE block.
Then, at the end, print out the entire array:
gawk 'BEGINFILE{ ++col } # New file, increment column number
{ X[FNR SEP col]=$0; rows=FNR } # Save datum into array X, indexed by current record number and col
END { for(r=1;r<=rows;r++){
comma=","
for(c=1;c<=col;c++){
if(c==col)comma=""
printf("%s%s",X[r SEP c],comma)
}
printf("\n")
}
}' chirps*
SEP is just an unused character that makes a separator between indices. I am using gawk because BEGINFILE is useful for incrementing the column number.
Save the above in your HOME directory as merge. Then start a Terminal and, just once, make it executable with the command:
chmod +x merge
Now change directory to where your chirps are with a command like:
cd subdirectory/where/chirps/are
Now you can run the script with:
$HOME/merge
The output will rush past on the screen. If you want it in a file, use:
$HOME/merge > merged.csv

Answer 2

First make one file without pasting and change that file into a oneliner with tr:
cat */chirps_*.csv | tr "\n" "," > long.csv

Answer 3

If the goal is a file with 100,000 lines and 500 columns then something like this should work:
paste -d, chirps_*.csv > chirps_500_merge.csv
Additional code can be used to sort the chirps_... input files into any desired order before pasteing.

Answer 4

The error comes from ulimit, from man ulimit:
-n or --file-descriptor-count The maximum number of open file descriptors
On my system ulimit -n returns 1024.
Happily we can paste the paste output, so we can chain it.
find . -type f -name 'file_*.csv' |
sort |
xargs -n$(ulimit -n) sh -c '
tmp=$(mktemp);
paste -d, "$#" >$tmp;
echo $tmp
' -- |
xargs sh -c '
paste -d, "$#"
rm "$#"
' --
Don't parse ls output
Once we moved from parsing ls output to good find, we find all files and sort them.
the first xargs takes 1024 files at a time, creates temporary file, pastes the output into temporary and outputs the temporary file filename
The second xargs does the same with temporary files, but also removes all the temporaries
As the count of files would be 100*500=500000 which is smaller then 1024*1024 we can get away with one pass.
Tested against test data generated with:
seq 1 2000 |
xargs -P0 -n1 -t sh -c '
seq 1 1000 |
sed "s/^/ $RANDOM/" \
>"file_$(date --date="-${1}days" +%Y%m%d).csv"
' --
The problem seems to be much like foldl with maximum size of chunk to fold in one pass. Basically we want paste -d, <(paste -d, <(paste -d, <1024 files>) <1023 files>) <rest of files> that runs kind-of-recursively. With a little fun I came up with the following:
func() {
paste -d, "$#"
}
files=()
tmpfilecreated=0
# read filenames...c
while IFS= read -r line; do
files+=("$line")
# if the limit of 1024 files is reached
if ((${#files[#]} == 1024)); then
tmp=$(mktemp)
func "${files[#]}" >"$tmp"
# remove the last tmp file
if ((tmpfilecreated)); then
rm "${files[0]}"
fi
tmpfilecreated=1
# start with fresh files list
# with only the tmp file
files=("$tmp")
fi
done
func "${files[#]}"
# remember to clear tmp file!
if ((tmpfilecreated)); then
rm "${files[0]}"
fi
I guess readarray/mapfile could be faster, and result in a bit clearer code:
func() {
paste -d, "$#"
}
tmp=()
tmpfilecreated=0
while readarray -t -n1023 files && ((${#files[#]})); do
tmp=("$(mktemp)")
func "${tmp[#]}" "${files[#]}" >"$tmp"
if ((tmpfilecreated)); then
rm "${files[0]}"
fi
tmpfilecreated=1
done
func "${tmp[#]}" "${files[#]}"
if ((tmpfilecreated)); then
rm "${files[0]}"
fi
PS. I want to merge all the csv files into a single csv in chronological order. Wouldn't that be just cut? Right now each column represents one day.

Answer 5

You can try this Perl-one liner. It will work for any number of files matching *.csv under a directory
$ ls -1 *csv
file_1.csv
file_2.csv
file_3.csv
$ cat file_1.csv
1
2
3
$ cat file_2.csv
4
5
6
$ cat file_3.csv
7
8
9
$ perl -e ' BEGIN { while($f=glob("*.csv")) { $i=0;open($FH,"<$f"); while(<$FH>){ chomp;#t=#{$kv{$i}}; push(#t,$_);$kv{$i++}=[#t];}} print join(",",#{$kv{$_}})."\n" for(0..$i) } ' <
1,4,7
2,5,8
3,6,9
$