Combining multiple =~ s/

hennesse has asked for the wisdom of the Perl Monks concerning the following question:

Replies are listed 'Best First'.
Re: Combining multiple =~ s/ by salva (Canon) on Mar 07, 2021 at 19:05 UTC
Let Perl generate the regular expression from a hash containing the list of words and their replacements: `my %trans = (zero => 0, one => 1, two => 2, three => 3); my $words_re = join "\|", map quotemeta, keys %trans; $string =~ s/($words_re)/$trans{lc $1}/ig;` [download]	[reply] [d/l]
Re^2: Combining multiple =~ s/ by Corion (Patriarch) on Mar 07, 2021 at 19:10 UTC
You want to sort the words in the regular expression by descending length :) Or alternatively, use `\b` to match only whole words: `my %trans = (zero => 0, one => 1, ones => 11, two => 2, three => 3); my $words_re = join "\|", map quotemeta, sort { length($b) <=> length($ +a) } keys %trans; $string =~ s/\b($words_re)\b/$trans{lc $1}/ig;` [download] Otherwise, it could be that `one` matches before `ones`. Update: choroba spotted that the `length($b)` and `length($a)` were missing	[reply] [d/l] [select]
Re^3: Combining multiple =~ s/ by no longer just digit (Beadle) on Mar 10, 2021 at 06:54 UTC
It might be worth pointing out the CPAN module Data::Munge which has a function `list2re` to make a list-matching regex. This line is basically what you need: `my $re = join '\|', map quotemeta, sort {length $b <=> length $a \|\| +$a cmp $b } @_;` [download] There is also this, which is what I use.	[reply] [d/l] [select]
Re^2: Combining multiple =~ s/ by LanX (Saint) on Mar 07, 2021 at 20:44 UTC
Nitpick! :) This is semantically not the same, even if sorted. Tho probably what the OP wants, if it's really about numbers Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery}	[reply]
Re: Combining multiple =~ s/ by haukex (Archbishop) on Mar 07, 2021 at 20:24 UTC
Building Regex Alternations Dynamically shows how to do this.	[reply]
Re: Combining multiple =~ s/ by kcott (Archbishop) on Mar 08, 2021 at 06:45 UTC
G'day Dave, "This is cool if there are just a few, but not if a lot. Is there a more compact way to do this?" If there are "just a few", you can still make this more compact by chaining non-destructive substitution operations. That was introduced in `5.14`; see "perl5140delta: Non-destructive substitution". `$string =~ s/zero/0/igr =~ s/one/1/igr =~ s/two/2/igr =~ s/three/3/igr +;` [download] If "a lot", then earlier advice regarding a pattern with alternation and a replacement using a lookup table, would be my choice. In either case, use a boundary assertion as already described; avoid `s/bones/b1s/` and similar mistakes. — Ken	[reply] [d/l] [select]
Re: Combining multiple =~ s/ by LanX (Saint) on Mar 07, 2021 at 20:50 UTC
the exact translation is in a loop `DB<1> @a= qw/zero one two three four/ DB<2> $pi ='three . one four one' DB<3> $pi =~ s/$a[$_]/$_/ig for 0..4 DB<4> p $pi 3 . 1 4 1 DB<5>` [download] Please note that replacing all the "one" before replacing "two" might change the input in unexpected ways. It really depends if you are really replacing numbers ... Cheers Rolf _{(addicted to the Perl Programming Language :) Wikisyntax for the Monastery}	[reply] [d/l]
Re: Combining multiple =~ s/ by rsFalse (Chaplain) on Mar 11, 2021 at 08:28 UTC
One more way. `#!/usr/bin/perl use warnings; use strict; my %huge_trans = ( zero => 0, one => 1, ones => 11, two => 2, three => 3, # ... million => 1_000_000, ); my $rx_word = qr/\b\w+\b/i; print <> =~ s!($rx_word)! $huge_trans{lc $1} // $1 !ger;` [download] Imagine you have huge dictionary for translation. Then this way should work faster. Previously suggested way with alternations should also work fast, because Trie-optimization should kick-in. Upd. BTW, it is not about "combining regex'es", it's only a use of a hash. All words, i.e. hash keys, must be simple words, not regexes.	[reply] [d/l]