Re: Is there another way to define a regex?
On 2016-01-17 Tom Browder wrote:
> My question: Is there a way to have Perl 6 do the required escaping
> for the regex programmatically, i.e., turn this:
>
> my $str = '/home/usr/.cpan';
>
> into this:
>
> my regex dirs {
> \/home\/usr\/\.cpan
> }
>
> automatically?
Yes! And it's also simpler than in Perl 5. I Perl 5, you would have to
do something like:
my $dirs = qr{\Q$str};
but in Perl 6 you just do:
my $dirs = regex { $str };
because the other behaviour, to interpret the contents of $str as
another regex to match, has more explicit syntax:
regex { <$another_regex> }
For your initial use-case there is also another shortcut: an array is
interpreted as an alternation, so you could write:
my @dirs = < /home/user/.cpan /home/tbrowde/.emacs >;
my $regex = regex { @dirs };
and it would do what your Perl 5 example does.
If you want to be more restrictive, you can anchor the alternation:
my $regex = regex { ^ @dirs };
Here is a complete program:
use v6;
my @dirs = < foo bar baz >;
my $matcher = regex { ^ @dirs $ };
for $*IN.lines -> $line {
if $line ~~ $matcher {
say "<$line> matched";
}
else {
say "<$line> didn't match";
}
}
--
Dakkar -
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Re: Is there another way to define a regex?
Curious but are the non capturing groups necessary?
On 17 Jan 2016 11:35 p.m., "Tom Browder" wrote:
> On Sun, Jan 17, 2016 at 3:03 PM, Moritz Lenz wrote:
> > On 01/17/2016 06:07 PM, Tom Browder wrote:
> ...
> >> # regex of dirs to ignore
> >> my regex dirs {
> >> \/home\/user\/\.cpan |
> >> \/home\/tbrowde\/\.emacs
> >> }
> >
> > Better written as
> >
> > my regex dirs {
> >| '/home/user/.cpan'
> >| '/home/tbowde/.emacs'
> > }
> >
> > Yes, quoting does now work in regexes too. Cool, right? :-)
>
> Yes, very cool! I have decided to use that format and it does ease
> adding or modifying entries.
>
> My final regex looks like this (shortened to only a couple of entries):
>
> my regex dirs {
> ^
> \s*
> [# <= non-capture grouping
>| '/home/user/.cpan'
>| '/home/tbowde/.emacs'
># more entries...
> ]
> }
>
> used like this:
>
> next LINE if $line ~~ //;
>
> Thanks, Moritz!
>
> Best,
>
> -Tom
>
Re: Is there another way to define a regex?
On Sunday, January 17, 2016, Nelo Onyiah wrote: > Curious but are the non capturing groups necessary? > I don't know the answer for the Perl 6 construct but I think the grouping is required in the Perl 5 equivalent. But I certainly may be wrong. Now that you mention it, I think you are probably correct. Thanks for pointing that out. -Tom
Re: Is there another way to define a regex?
On Sun, Jan 17, 2016 at 3:03 PM, Moritz Lenz wrote:
> On 01/17/2016 06:07 PM, Tom Browder wrote:
...
>> # regex of dirs to ignore
>> my regex dirs {
>> \/home\/user\/\.cpan |
>> \/home\/tbrowde\/\.emacs
>> }
>
> Better written as
>
> my regex dirs {
>| '/home/user/.cpan'
>| '/home/tbowde/.emacs'
> }
>
> Yes, quoting does now work in regexes too. Cool, right? :-)
Yes, very cool! I have decided to use that format and it does ease
adding or modifying entries.
My final regex looks like this (shortened to only a couple of entries):
my regex dirs {
^
\s*
[# <= non-capture grouping
| '/home/user/.cpan'
| '/home/tbowde/.emacs'
# more entries...
]
}
used like this:
next LINE if $line ~~ //;
Thanks, Moritz!
Best,
-Tom
Re: Is there another way to define a regex?
On Sun, Jan 17, 2016 at 2:10 PM, Tom Browder wrote:
> On Sun, Jan 17, 2016 at 1:55 PM, Bruce Gray wrote:
>> On Jan 17, 2016, at 11:07 AM, Tom Browder wrote:
>>
>>> I'm trying to write all new Perl code in Perl 6. One thing I need is
>>> the equivalent of the Perl 5 qr// and, following Perl 6 docs, I've
> ...
>
> I'll try all that, Bruce. Thanks!
Hm, I'm not getting good results (> 20 sec vs 10 sec). I realize I
may have oversimplified what I'm trying to do. My incoming data lines
are complete file names. I want to ignore file names whose path meets
certain partial paths. So my working Perl 6 example, fully-escaped,
should be:
my regex dirs_to_ignore {
^ \s* [ # non-capture grouping
\/home\/user\/\.cpan |
\/home\/user\/some\-dir
]
}
And it runs very fast and accurately. When I remove the backslashes
and use single quotes around the partial paths it also DOES work as
shown here!!
my regex dirs_to_ignore {
^ \s* [ # non-capture grouping
'/home/user/.cpan' |
'/home/user/some-dir'
]
}
I probably made some subtle error previously. Sorry for the wasted bandwidth.
I also tried again auto-building the regex but with no final success.
I could get the string exactly as in the second example bout I
couldn't get it to compile or to be used as a regex as far as I could
tell by timing and output comparison.
Best,
-Tom
Re: Is there another way to define a regex?
On 01/17/2016 06:07 PM, Tom Browder wrote:
> I'm trying to write all new Perl code in Perl 6. One thing I need is
> the equivalent of the Perl 5 qr// and, following Perl 6 docs, I've
> come to use something like this (I'm trying to ignore certain
> directories):
>
> # regex of dirs to ignore
> my regex dirs {
> \/home\/user\/\.cpan |
> \/home\/tbrowde\/\.emacs
> }
Better written as
my regex dirs {
| '/home/user/.cpan'
| '/home/tbowde/.emacs'
}
Yes, quoting does now work in regexes too. Cool, right? :-)
(The leading | is ignored, it just allows you to format the alternation
more consistently)
> for "dirlist.txt".IO.lines -> $line {
> # ignore certain dirs
> if $line ~~ m{} {
> next;
> }
> }
>
> My question: Is there a way to have Perl 6 do the required escaping
> for the regex programmatically, i.e., turn this:
>
> my $str = '/home/usr/.cpan';
>
> into this:
>
> my regex dirs {
> \/home\/usr\/\.cpan
> }
>
> automatically?
Even better: No need to escape anymore. If you use $str in a regex, and
it actually contains a Str, it is always taken to match literally, so
my $dir1 = '/home/user/.cpan';
my $dir2 = '/home/tbowde/.emacs';
my regex ignore_dirs { $dir1 | $dir2 }
does what you want.
If you want the interpolated string to be treated as a regex, you have
to use it as my regex dirs { <$dir1> }.
Cheers,
Moritz
Re: Is there another way to define a regex?
On Sun, Jan 17, 2016 at 1:55 PM, Bruce Gray wrote: > On Jan 17, 2016, at 11:07 AM, Tom Browder wrote: > >> I'm trying to write all new Perl code in Perl 6. One thing I need is >> the equivalent of the Perl 5 qr// and, following Perl 6 docs, I've ... I'll try all that, Bruce. Thanks! -Tom
Re: Is there another way to define a regex?
On Jan 17, 2016, at 11:07 AM, Tom Browder wrote:
> I'm trying to write all new Perl code in Perl 6. One thing I need is
> the equivalent of the Perl 5 qr// and, following Perl 6 docs, I've
> come to use something like this (I'm trying to ignore certain
> directories):
>
> # regex of dirs to ignore
> my regex dirs {
> \/home\/user\/\.cpan |
> \/home\/tbrowde\/\.emacs
> }
>
> for "dirlist.txt".IO.lines -> $line {
> # ignore certain dirs
> if $line ~~ m{} {
>next;
> }
> }
>
> My question: Is there a way to have Perl 6 do the required escaping
> for the regex programmatically, i.e., turn this:
>
> my $str = '/home/usr/.cpan';
>
> into this:
>
> my regex dirs {
> \/home\/usr\/\.cpan
> }
>
> automatically?
I think that your regex needs anchors to do what you want;
your current regex will also exclude `/home/user/.cpanabcedfg`, for example.
This is how I do it in Perl 5 (when using regexes instead of a hash):
my $dirs = join '|', map { quotemeta } qw(
/home/user/.cpan
/home/tbrowde/.emacs
);
my $dirs_re = qr/^(?:$dirs)$/;
So, `quotemeta` is what you are looking for.
Except that https://design.perl6.org/S29.html#Obsolete_Functions says:
quotemeta
Because regex escaping metacharacters can easily be solved by quotes
("Simplified lexical parsing of patterns" in S05),
and variables are not automatically interpolated
("Variable (non-)interpolation" in S05),
quotemeta is no longer needed.
http://design.perl6.org/S05.html#Simplified_lexical_parsing_of_patterns
http://design.perl6.org/S05.html#Variable_%28non-%29interpolation
Summary of `Variable_%28non-%29interpolation`:
In Perl 5, /$var/ deposits the contents of `$var` into the regex,
just like of you had typed them in the source code.
In Perl 6, /$var/ runs quotemeta on the contents of `$var`.
If you want the Perl 5 behavior, then you write it as /<$var>/.
Also, if you embed an array in a regex, it automatically treats it as `|`
alternatives.
So cool!
# Plain text, *not* regex!
my @dirs_to_skip = <
/home/user/.cpan
/home/tbrowde/.emacs
> ;
...
next if $line ~~ / ^ @dirs_to_skip $ /;
Or, precompiled:
my @dirs_to_skip = <
/home/user/.cpan
/home/tbrowde/.emacs
> ;
my $dir_re = / ^ @dirs_to_skip $ /;
...
next if $line ~~ $dir_re;
If you know that your exclusion list is always literal,
you can leave the regex engine out of the process,
and just do a hash lookup
(which is even easier in Perl 6 with `set`, which sets the values all to
`True`):
my %dirs_to_skip = set <
/home/user/.cpan
/home/tbrowde/.emacs
> ;
...
next if %dirs_to_skip{$line};
--
Hope this helps,
Bruce Gray (Util of PerlMonks)
Is there another way to define a regex?
I'm trying to write all new Perl code in Perl 6. One thing I need is
the equivalent of the Perl 5 qr// and, following Perl 6 docs, I've
come to use something like this (I'm trying to ignore certain
directories):
# regex of dirs to ignore
my regex dirs {
\/home\/user\/\.cpan |
\/home\/tbrowde\/\.emacs
}
for "dirlist.txt".IO.lines -> $line {
# ignore certain dirs
if $line ~~ m{} {
next;
}
}
My question: Is there a way to have Perl 6 do the required escaping
for the regex programmatically, i.e., turn this:
my $str = '/home/usr/.cpan';
into this:
my regex dirs {
\/home\/usr\/\.cpan
}
automatically?
Thanks.
Best regards,
-Tom
