perl - replace every nth (and multiples) occurrences of a character with another character

Question

Does anyone know any unix commands/perl script that would insert a specific character (that can be entered as either hex (ie 7C) or as the actual character (ie |)) in the po

Answer 1

use File::Slurp qw(read_file);
my ($from, $to, $every, $fname) = @ARGV;
my $counter = 0;
my $in = read_file $fname;
my $out = $in;
# copy is important because pos magic attached to $in resets with substr
while ($in =~ /\Q$from/gms) {
    $counter++;
    substr $out, pos($in)-1, length($from), $to unless $counter % $every;
};
print $out;

If the $from and $to parameters have different length, you still need to mess a bit with the second parameter of substr to make it work correctly.

Answer 2

Small perl hack to solve the problem. Using the index function to find the commas, modulus to replace the right one, and substr to perform the replacement.

use strict;
use warnings;

while (<>) {
    my $x=index($_,","); 
    my $i = 0; 
    while ($x != -1) {
        $i++; 
        unless ($i % 3) { 
            $_ = substr($_,0,$x) ."|". substr($_,$x+1); 
        }
        $x = index($_,",",$x + 1) 
    } 
    print;
}

Run with perl script.pl file.csv.

Note: You can place the declaration my $i before the while(<>) loop in order to do a global count, instead of a separate count for each line. Not quite sure I understood your question in that regard.

Answer 3

This processes the input file one line at a time (no slurping :)
For hex input, just pass '\x7C' or whatever, as $1

#!/bin/bash  

b="${1:-,}"                             # the "before" field delimiter 
n="${2:-3}"                             # the number of fields in a group
a="${3:-|}"; [[ $a == [\|] ]] && a='\|' # the "after" group delimiter

sed -nr "x;G; /(([^$b]+$b){$((n-1))}[^$b]+)$b/{s//\1$a/g}
         s/.*\n//; h; /.*$a/{s///; x}; p" input_file

---

Here it is again, with some comments.

sed -nr "x;G    # pat = hold + pat
  /(([^$b]+$b){$((n-1))}[^$b]+)$b/{s//\1$a/g}
  s/.*\n//      # del fields from prev line
  h             # hold = mod*\n
  /.*$a/{ s///  #  pat = unmodified
          x     # hold = unmodified, pat = mod*\n
        }            
  p             # print line"  input_file

Answer 4

I have an idea in bash script :

perl -pe 's/,/(++$n % 3 == 0) ? "|" : $&/ge'  data.txt

That will do the trick.

Answer 5

How about a nice, simple awk one-liner?

awk -v RS=, '{ORS=(++i%3?",":"|");print}' file.csv

One minor bug just occurred to me: it will print a , or | as the very last character. To avoid this, we need to alter it slightly:

awk -v RS=, '{ORS=(++i%3?",":"|");print}END{print ""}' file.csv | sed '$d'

Answer 6

# Get params and create part of the regex.
my $delim   = "\\" . shift;
my $n       = shift;
my $repl    = shift;
my $wild    = '.*?';
my $pattern = ($wild . $delim) x ($n - 1);

# Slurp.
$/       = undef;
my $text = <>;

# Replace and print.
$text =~ s/($pattern$wild)$delim/$1$repl/sg;
print $text;