Re: Perl to Ruby: regex captures to assignment.

Try this:

w, n, d = frac_str =~ %r{/} \
? frac_str.match(%r{(?:(\S+) )*?(\S+)/(\S+)}).captures \
: [1, frac_str, 0]

I've use %r{} to quote regex, so no need to escape for /


On Thu, Dec 20, 2012 at 8:45 AM, Derrick B. <[e-mail]> wrote:

> Hello all,
>
> I am converting a Perl program to Ruby in order to learn Ruby.  There is
> an expression that takes a string, ("1 3/4" or "5", for example)
> determines what kind of fraction was entered, and assigns variables (w,
> n, d), appropriately.  If a whole number was entered, then it assigns
> the variables differently, of course.
>
> The expression first checks for the presence of a slash ("/") to
> determine if it is a fraction or whole number.  Then, using the
> conditional operator, assigns the variables, appropriately.
>
> Here are the two expressions:
>
> Perl:
>
>     my ($w, $n, $d) = ( $frac_str =~ /\// )
>             ? $frac_str =~ /(?:(\S+) )??(\S+)\/(\S+)/
>             : ( 1, $frac_str, 0 );
>
> Ruby:
>
>     w, n, d = frac_str.match(/\//) \
>         ? frac_str.match(/(?:(\S+) )*?(\S+)\/(\S+)/).captures \
>         : 0, frac_str, 1
>     puts "w: #{w}, n: #{n}, d: #{d}"  # check assignment
>
>
> I used irb to test the regex and it works just fine:
>
> exp = Regexp.new(/(?:(\S+) )*?(\S+)\/(\S+)/)
> => /(?:(\S+) )*?(\S+)\/(\S+)/
>
> str = "1 3/4"
> => "1 3/4"
>
> n, w, d = str.match(exp).captures
> => ["1", "3", "4"]
>
> w
> => "3"
>
> n
> => "1"
>
> d
> => "4"
>
> But, the above version outputs:
>
> w: 134, n: 1 3/4, d: 1
>
> It appears to recurse over the regex and duplicate the captures.  But,
> since I am new to this language, I know I can use an 'if..else'
> statement, but since I am coming from Perl, I figured that a similar
> expression would "just work"!  :)
>
> Thanks for your time and input,
>
> Derrick
>
> --
> Posted via http://www.ruby-forum.com/.
>
>


-- 
Regards,
Wei Feng

03 9005 3441 | M 0413 658 250 | [e-mail] | http://wei.feng.id.au