How do we flatten or stringify Match (or else) object to be string data type (esp. in multitude ie. as array elements)? e.g.
'foobar' ~~ m{ (foo) };
say $0.WHAT;
my $foo = $0;
say $foo.WHAT
(Match)
(Match)
How to end up with (Str)?
~ is the Str contextualizer:
'foobar' ~~ m{ (foo) };
say ~$0
will directly coerce it to a Str. You can use that if you have many matches, i. e.:
'foobar' ~~ m{ (f)(o)(o) };
say $/.map: ~*; # (f o o)
Just treat the objects as if they were strings.
If you apply a string operation to a value/object Raku will almost always just automatically coerce it to a string.
String operations include functions such as print and put, operators such as infix eq and ~ concatenation, methods such as .starts-with or .chop, interpolation such as "A string containing a $variable", and dedicated coercers such as .Str and Str(...).
A Match object contains an overall match. Any "children" (sub-matches) are just captures of substrings of that overall match. So there's no need to flatten anything because you can just deal with the single overall match.
A list of Match objects is a list. And a list is itself an object. If you apply a string operation to a list, you get the elements of the list stringified with a space between each element.
So:
'foobar' ~~ m{ (f) (o) (o) };
put $/; # foo
put $/ eq 'foo'; # True
put $/ ~ 'bar'; # foobar
put $/ .chop; # fo
put "[$/]"; # [foo]
put $/ .Str; # foo
my Str() $foo = $/;
say $foo.WHAT; # (Str)
put 'foofoo' ~~ m:g{ (f) (o) (o) }; # foo foo
The constructor for Str takes any Cool value as argument, including a regex Match object.
'foobar' ~~ m{ (foo) };
say $0.WHAT; # (Match)
say $0.Str.WHAT; # (Str)
Related
Should be very simple, but I can't cope with it.
I want to match exactly the same number of as as bs. So, the following
my $input = 'aaabbbb';
$input ~~ m:ex/ ... /;
should produce:
aaabbb
aabb
ab
UPD: The following variants don't work, perhaps because of the :ex bug , mentioned in #smls's answer (but more likely because I made some mistakes?):
> my $input = "aaabbbb";
> .put for $input ~~ m:ex/ (a) * (b) * <?{ +$0 == +$1 }> /;
Nil
> .put for $input ~~ m:ex/ (a) + (b) + <?{+$0 == +$1}> /;
Nil
This one, with :ov and ?, works:
> my $input = "aaabbbb";
> .put for $input ~~ m:ov/ (a)+ (b)+? <?{+$0 == +$1}> /;
aaabbb
aabb
ab
UPD2: The following solution works with :ex as well, but I had to do it without <?...> assertion.
> $input = 'aaabbbb'
> $input ~~ m:ex/ (a) + (b) + { put $/ if +$0 == +$1 } /;
aaabbb
aabb
ab
my $input = "aaabbbb";
say .Str for $input ~~ m:ov/ (a)+ b ** {+$0} /;
Output:
aaabbb
aabb
ab
It's supposed to work with :ex instead of :ov, too - but Rakudo bug #130711 currently prevents that.
my $input = "aaabbbb";
say .Str for $input ~~ m:ov/ a <~~>? b /;
Works with ex too
my $input = "aaabbbb";
say .Str for $input ~~ m:ex/ a <~~>? b /;
Upd: explanation
<~~> means call myself recursively see Extensible metasyntax. (It is not yet fully implemented.)
Following (longer, but maybe clearer) example works too:
my $input = "aaabbbb";
my token anbn { a <&anbn>? b}
say .Str for $input ~~ m:ex/ <&anbn> /;
Here I make a regex manually from Regex elements of an array.
my Regex #reg =
/ foo /,
/ bar /,
/ baz /,
/ pun /
;
my $r0 = #reg[0];
my $r1 = #reg[1];
my Regex $r = / 0 $r0 | 1 $r1 /;
"0foo_1barz" ~~ m:g/<$r>/;
say $/; # (｢0foo｣ ｢1bar｣)
How to do it with for #reg {...}?
If a variable contains a regex, you can use it without further ado inside another regex.
The second trick is to use an array variable inside a regex, which is equivalent to the disjunction of the array elements:
my #reg =
/foo/,
/bar/,
/baz/,
/pun/
;
my #transformed = #reg.kv.map(-> $i, $rx { rx/ $i $rx /});
my #match = "0foo_1barz" ~~ m:g/ #transformed /;
.say for #match;
my #reg =
/foo/,
/bar/,
/baz/,
/pun/
;
my $i = 0;
my $reg = #reg
.map({ $_ = .perl; $_.substr(1, $_.chars - 2); })
.map({ "{$i++}{$_}" })
.join('|');
my #match = "foo", "0foo_1barz" ~~ m:g/(<{$reg}>) /;
say #match[1][0].Str;
say #match[1][1].Str;
# 0foo
# 2baz
See the docs
Edit: Actually read the docs myself. Changed implicit eval to $() construct.
Edit: Rewrote answer to something that actually works
Edit: Changed answer to a terrible, terrible hack
I want to make all the consonants in a word uppercase:
> my $word = 'camelia'
camelia
> $word ~~ s:g/<-[aeiou]>/{$/.uc}/
(｢c｣ ｢m｣ ｢l｣)
> $word
CaMeLia
To make the code more general, I store the list of all the consonants in a string variable
my $vowels = 'aeiou';
or in an array
my #vowels = $vowels.comb;
How to solve the original problem with $vowels or #vowels variables?
Maybe the trans method would be more appropriate than the subst sub or operator.
Try this:
my $word = "camelia";
my #consonants = keys ("a".."z") (-) <a e i o u>;
say $word.trans(#consonants => #consonants>>.uc);
# => CaMeLia
With the help of moritz's explanation, here is the solution:
my constant $vowels = 'aeiou';
my regex consonants {
<{
"<-[$vowels]>"
}>
}
my $word = 'camelia';
$word ~~ s:g/<consonants>/{$/.uc}/;
say $word; # CaMeLia
You can use <!before …> along with <{…}>, and . to actually capture the character.
my $word = 'camelia';
$word ~~ s:g/
<!before # negated lookahead
<{ # use result as Regex code
$vowel.comb # the vowels as individual characters
}>
>
. # any character (that doesn't match the lookahead)
/{$/.uc}/;
say $word; # CaMeLia
You can do away with the <{…}> with #vowels
I think it is also important to realize you can use .subst
my $word = 'camelia';
say $word.subst( :g, /<!before #vowels>./, *.uc ); # CaMeLia
say $word; # camelia
I would recommend storing the regex in a variable instead.
my $word = 'camelia'
my $vowel-regex = /<-[aeiou]>/;
say $word.subst( :g, $vowel-regex, *.uc ); # CaMeLia
$word ~~ s:g/<$vowel-regex>/{$/.uc}/;
say $word # CaMeLia
Is it possible to define nested regexes in arbitrary sequence?
The following program works as expected:
my regex letter { <[a b]> }
my regex word { <letter> + }
my $string = 'abab';
$string ~~ &word;
put $/; # abab
If I swap the first two lines, compiler produces an error.
Is there a way to override this restriction (without using grammars)?
You can put the regex in a variable you declare up front but later set:
my $letter;
my regex word { <$letter> + }
$letter = regex { <[a b]> }
my $string = 'abab';
$string ~~ &word;
put $/; # abab
I'm trying to replace a word in a string. The word is stored in a variable so naturally I do this:
$sentence = "hi this is me";
$foo=~ m/is (.*)/;
$foo = $1;
$sentence =~ s/$foo/you/;
print $newsentence;
But this doesn't work.
Any idea on how to solve this? Why this happens?
Perl lets you interpolate a string into a regular expression, as many of the answers have already shown. After that string interpolation, the result has to be a valid regex.
In your original try, you used the match operator, m//, that immediately tries to perform a match. You could have used the regular expression quoting operator in it's place:
$foo = qr/me/;
You can either bind to that directory or interpolate it:
$string =~ $foo;
$string =~ s/$foo/replacement/;
You can read more about qr// in Regexp Quote-Like Operators in perlop.
You have to replace the same variable, otherwise $newsentence is not set and Perl doesn't know what to replace:
$sentence = "hi this is me";
$foo = "me";
$sentence =~ s/$foo/you/;
print $sentence;
If you want to keep $sentence with its previous value, you can copy $sentence into $newsentence and perform the substitution, that will be saved into $newsentence:
$sentence = "hi this is me";
$foo = "me";
$newsentence = $sentence;
$newsentence =~ s/$foo/you/;
print $newsentence;
You first need to copy $sentence to $newsentence.
$sentence = "hi this is me";
$foo = "me";
$newsentence = $sentence;
$newsentence =~ s/$foo/you/;
print $newsentence;
Even for small scripts, please 'use strict' and 'use warnings'. Your code snippet uses $foo and $newsentence without initialising them, and 'strict' would have caught this. Remember that '=~' is for matching and substitution, not assignment. Also be aware that regexes in Perl aren't word-bounded by default, so the example expression you've got will set $1 to 'is me', the 'is' having matched the tail of 'this'.
Assuming you're trying to turn the string from 'hi this is me' to 'hi this is you', you'll need something like this:
my $sentence = "hi this is me";
$sentence =~ s/\bme$/\byou$/;
print $sentence, "\n";
In the regex, '\b' is a word boundary, and '$' is end-of-line. Just doing 's/me/you/' will also work in your example, but would probably have unintended effects if you had a string like 'this is merry old me', which would become 'this is yourry old me'.