Hi Shifus, I was trying to parse a csv and replace cell content according another text file. I was using this: 

 
# This script was excerpted from http://stackoverflow.com/questions/11
+678939/replace-text-based-on-a-dictionary
use strict;
use warnings;
use Text::CSV;

open my $fh, '<', 'kegg_pathway_title.txt' or die $!;
my %dict =  map { chomp; split ' ', $_, 2 } <$fh>;
my $re = join '|', keys %dict;

open $fh, '<', 'Orthogroups_3.csv' or die $!;

while (<$fh>) {
  next if $. < 2;
  s/($re)/$dict{$1}/g;
  print;
}

[download]

It gives me near expected output but changed the dimension of the original file. Now I can't load the parsed csv in R to do other stuff. Is there anyway to replace cell content of csv without changing dimension? 

 > grpsTbl <- read.csv("Orthogroups_3.csv", header=T, sep = "\t", row.
+names = 1, stringsAsFactors=F)
> dim(grpsTbl)
[1] 5791   13

[download]
 > grpsTbl <- read.csv("parsed_with_perl.csv", header=T, sep = "\t", r
+ow.names = 1, stringsAsFactors=F)
Error in read.table(file = file, header = header, sep = sep, quote = q
+uote,  : 
  more columns than column names

[download]

My orthogroups_3.csv contain multicolumn, each cell may contain 0/1/many values(name). I want exact same csv just values in each cell(names) will be replaced according do the 1st file. Best Regards Zillur

In reply to Parsing csv without changing dimension of original file by zillur

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