Hello, let us solve the text-processing Task 1 from Week 26 of the Perl Weekly Challenge. The task, after deciphering the original phrasing, is:
Take the two words and count all letters in the second word, which are present in the first word.
I have two solutions here, and I believe there is potential to make them shorter.
1
In the first solution, a set $alphabet is created out of the letters from the first word $pattern. Then you also split the second word into letters ($test.comb ) and filter that list so that only the letters from the $alphabet can escape. And print the number of such cases (.elems).
my ($pattern, $test) = @*ARGS;
my $alphabet = set($pattern.comb);
say $test.comb.grep({$alphabet{$_}:exists}).elems;
Run the program and pass two words to it:
$ raku ch-1.raku chancellor chocolate 8
2
In my second solution, no separate alphabet is needed; we scan the letters in the second word (for $test.comb) and try to find them in the first word ($pattern.match($_)). If it is found, then (&&) the counter is incremented: $count++.
my ($pattern, $test) = @*ARGS; my $count = 0; $pattern.match($_) && $count++ for $test.comb; say $count;
This program has the same interface as the previous one:
$ raku ch-1a.raku chancellor chocolate 8
3
And when you have the program completed, you understand that you can merge them to get a proper concise one-line solution!
my ($pattern, $test) = @*ARGS;
say $test.comb.grep({$pattern.match($_)}).elems;
And that’s all for now.
→ GitHub repository
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