Bash regex finding and replacing
<3 digit number> <data>

<word>:<4 digit number>

020 Word
021 Word:0001
Replace with
020 021
021 0210001

#!/usr/bin/perl

$in= "input.txt";
$out= "output.txt";

# Buffer the whole file for replacing:
open(INFILE, $in);
@lines = <INFILE>;
open(INFILE, $in);

# Iterate through each line:
while(<INFILE>) {
  # If the line matches "word:number", replace all instances in the file
  if (/^(d{3}) (w+:)d{4}$/) {
    $num = $1; word = $2;
    s/$word/$num/ foreach @lines;
  }
}

open(OUTFILE, $out);
print OUTFILE foreach @lines;

#file name:t
kent$  cat t
020 Word
021 Word:0001

#first we find out the replacement, 021 in this case:
kent$  v=$(grep -oP "(d{3})(?= Word:d{4})" t|head -n1)

#do replace by sed:
kent$  sed -r "s/Word[:]?/$v/g" t
020 021
021 0210001

number=$(gawk --posix '/[0-9]{3} '${word}':[0-9]{4}/ { print $1; exit }' $file)

if [ "$number" != "" ]; then
    sed -r "s/${word}:?/${number}/" $file
fi

gawk '
    NR == FNR {
        if (match($2, /^([^:]+):[0-9][0-9][0-9][0-9]$/, a))
            repl[a[1] ":?"] = $1
        next
    }
    {
        for (word in repl)
            if ($2 ~ word) {
                sub(word, repl[word], $2)
                break
            }
        print
    }
' filename filename > new.file

v=`grep -m1 'Word[:][0-9][0-9][0-9][0-9]'< filename | awk '{print $1}'` &&
sed -i '' 's/Word[:]*/'$v'/g' filename

# sweep the file and make a lookup table variable

lookup=$(sed -nr 's/(.*) (.*:).*/21/p' <source_file |tr 'n' ' ')

# append the lookup to each line and substitute using a backreference
# N.B. remove the lookup whatever!

     sed -r "s/$/@@${lookup}/;
             s/^(... )(.*)$@@.*2:(S*).*/13/;
             s/^(... )(.*:)(.*)@@.*2(S*).*/143/;
             s/@@.*//" <source_file