Combining multiple files by year (column in data) into one line and add file name as first column in Perl or Python

I created fake intput data like this:

#!/usr/bin/perl use warnings; use strict; for my $id ('aaa' .. 'eod') { open my $out, '>', "file.$id" or die "file.$id: $!"; for my $year (1979 .. 2023) { for my $month (1 .. 12) { printf {$out} "%d %d 0 %f\n", $year, $month, rand() / (1 + rand); } } } 

And then processed them by the following Perl script:

#!/usr/bin/perl use warnings; use strict; use feature 'say'; my %by_year; for my $file (glob 'file.???') { my $id = substr $file, -3; open my $in, '<', $file or die "$file: $!"; while (my $line = <$in>) { my ($year, $month, undef, $precipitation) = split ' ', $line; die "Duplicate $id $year $month" if exists $by_year{$year}{$id}{$month}; $by_year{$year}{$id}{$month} = $precipitation; } } for my $year (keys %by_year) { open my $out, '>', "$year.out" or die "$year.out: $!"; for my $id (sort keys %{ $by_year{$year} }) { say {$out} join ' ', $id, map { $_, $by_year{$year}{$id}{$_} } 1 .. 12; } } 

You didn't show how exactly your files are named, so you'll have to adjust the line that does the glob as well as the one that extracts the id (using substr here).

Ahoy!

I was able to sort through it. The steps I followed were as follows.

1. A separate script to create 3000 files of the format "YEAR, MONTH, REMOVE THIRD COLUMN, PRECIPITATION"

2. The files created have a random name, an underscore, the order they were created, and a .txt extension i.e. 17920_2624.txt would be file number 2624

3. The precipitation in the randomly created files is a random number between 0-250, and a decimal between 0-999 i.e. 229.370

4. The main script will parse each of these 3000 files, and store the data in hash %filenameYears. The hash key is each filename, the hash value is a pointer to %allYearsInFile. %allYearsInFile will become an anonymous hash containing precipitation data for each year in the file

5. The anonymous hash key is each year, the anonymous hash value is the output string of data for that year in the condensed formatted you requested

6. If the precipitation data for any month is missing, substitute the missing value with -99.99

The scipt to create the output files looks like this...

#!/usr/bin/perl -w my $minimum = 1980; my $maximum = 2025; my $nfiles = shift or die("no command line arg"); my $count = 0; my @header = ("YEAR,", "MONTH,", "REMOVE THIRD COLUMN,", "PRECIPITATION"); my @lengthHeaders; #for printf table formatting for(@header){ $lengthHeaders[$count] = "%-" . length($_) . "s"; $count++; } for $count (1 .. $nfiles) { my $filename = int(rand(99999)); $filename .="_${count}.txt"; open my $out, '>', "$filename" or die "$filename: $!"; printf($out "@lengthHeaders\n",@header); for my $year ($minimum .. $maximum) { for my $month (1 .. 12) { printf($out "@lengthHeaders\n", $year, $month, ,0, int(rand(250)) . "." . int(rand(999))); } } } #run this command to print the files in the order they were created #perl -e 'print "$_\n" for(sort { ($a =~ /_(\d+)\.txt/)[0] <=> ($b =~ /_(\d+)\.txt/)[0] } @ARGV)' *.txt 

This script will generate however many files you specify. To create 3000 files, run the following command...

$ perl create.files.pl 3000 

Here is the code to parse the input files and put them in the condensed format you described...

#!/usr/bin/perl -w my @headers = ("IDENTIFICATION,", "YEAR,", "MONTH CODE,", "PRECIPITATION FOR THAT MONTH"); my @lengthHeaders; #find length of each header for printf table formatting my ($file,$filecount,$count) = ("",0,0); for(@headers){ $lengthHeaders[$count] = "%-" . length($_) . "s"; $count++; } my %filenameYears; #key is filename, value is a pointer to an anonymous hash containing data from all years in the file my %allYearsInFile; #key is year, value is output string of data for that year while(<>){ if($file ne $ARGV){ #filename being processed has just changed $filecount++; $file = $ARGV; #update new filename next; #skip header lines } my %line; @line{("year","month","remove","precipitation")} = split(/ +/); if(!defined($allYearsInFile{$line{year}})){ #start a new year value #$allYearsInFile{$line{year}} = $line{year}; $allYearsInFile{$line{year}} = sprintf("$lengthHeaders[1]", $line{year} . ","); } if( $line{precipitation} !~ /\d/){ #if precipitation value is missing, set value to -99.99 #DEBUG: warn "missing value $_"; $line{precipitation} = -99.99; } #$allYearsInFile{$line{year}} .= ", $line{month}, $line{precipitation}"; $allYearsInFile{$line{year}} .= sprintf( " %-4s%-8s", $line{month} . ",", $line{precipitation} . ","); if($line{month} == 12){ $allYearsInFile{$line{year}} =~ s/, *$//; #remove trailing comma } if(eof){ #when file ends, save hash and start a new one for the next file $filenameYears{$file} = {%allYearsInFile}; #save old hash as anonymous hash by filename %allYearsInFile = (); #empty old hash for next year } } printf("@lengthHeaders\n",@headers); #done processing files, print headers #DEBUG: print "Processed $filecount files\n"; for my $filename (sort keys %filenameYears){ #dereference hash and print output string in formatted printf table my $hashref = $filenameYears{$filename}; for my $year ( sort keys %$hashref ){ #print "$filename, $$hashref{$year}\n"; printf("@lengthHeaders[0..1]\n", $filename . ",", $$hashref{$year}); } } 

To parse the 3000 files and put them in the condensed format you described, run the following command...

$ perl parse.precipitation.files.pl *.txt IDENTIFICATION, YEAR, MONTH CODE, PRECIPITATION FOR THAT MONTH 84406_1.txt, 1980, 1, 187.288, 2, 22.298, 3, 175.23, 4, 41.606, 5, 104.842, 6, 176.260, 7, 207.896, 8, 143.67, 9, 57.9, 10, 69.99, 11, 146.85, 12, 49.121 84406_1.txt, 1981, 1, 128.77, 2, 242.826, 3, 49.836, 4, 115.318, 5, 79.676, 6, 2.585, 7, 109.714, 8, 100.613, 9, 123.566, 10, 218.599, 11, 115.717, 12, 76.219 84406_1.txt, 1982, 1, 227.123, 2, 155.287, 3, 95.521, 4, 17.647, 5, 176.328, 6, 95.766, 7, 106.289, 8, 90.45, 9, 93.676, 10, 142.85, 11, 141.379, 12, 109.357 <cut> 

This will parse all 3000 files sequentially and print the output in the format you requested. If you want to save this output in its own file, run the following command...

$ perl parse.precipitation.files.pl *.txt > year_mly.txt 

For all 3000 files it should take around 5 seconds to finish.

$ time perl parse.precipitation.files.pl *.txt > year_mly.txt real 0m4.345s user 0m4.282s sys 0m0.061s 

To separate this large file into files by year, i.e. one file containing all readings from all files for the year 1980, you can run the following script...

#!/usr/bin/perl -w my $headerLine = 1; my @headers = ("IDENTIFICATION", "YEAR", "MONTH CODE", "PRECIPITATION FOR THAT MONTH"); my @lengthHeaders; #find length of each header for printf table formatting my ($file,$filecount,$count) = ("",0,0); for(@headers){ $lengthHeaders[$count] = "%-" . length($_) . "s"; $count++; } my %allFilesByYear; while(<>){ if($headerLine){ $headerLine = 0; next; } my %line; @line{@headers} = split(/, +/); push(@{$allFilesByYear{$line{YEAR}}},$_); #store each line in a hash where the key is each year, and the value is the row of data. } for $k (sort keys(%allFilesByYear)){ #dereference hash my $arrayref = $allFilesByYear{$k}; open my $out,'>',"${k}_mly.txt" or die "$!"; #create new file for each year for( @{$allFilesByYear{$k}} ){ print $out "$_"; #output data for each year } } 

Run this script with the following command...

$ perl files.by.year.pl year_mly.txt 

This will automatically separate all the data into smaller files based on the year. So all the readings from 1980 will be in the file 1980_mly.txt and so on for each year. You can also accomplish this manually using something like grep. To see all data for the year 1980 manually, you could run something like this...

more year_mly.txt | grep -i ' 1980,' 

And to put this data in a file you could run something like this...

more year_mly.txt | grep -i ' 1980,' > data_from_1980.txt 

That should be exactly what you are looking for. If the requirements arent all met just let me know what you need with some sample data.

Good Luck!