Ah, nice to see someone else tackling the problems. Assuming the tests
pass (I can't check this at work) and you have a commit bit, check it
in!
Cheers,
Ovid
--- gabriele renzi <[EMAIL PROTECTED]> wrote:
> Hi everyone, and happy new year!
>
> I'm an almost complete newbie to Perl6 and I'm not that good at Perl5
>
> either, but I thought playing with these problems could be fun, so I
> tried to solve #28.
>
> My solution is attached, it seems to work, but I'd like to know from
> people more expert than me if there could be a better solution.
>
> The first part seems reasonable, the second could probably be a
> one-liner.
> Probably I could replace usage of hash+for with gather/take and
> nested
> lists, but I don't think this makes the solution cleaner.
>
> Thanks in advance for any suggestion.
> > use v6-alpha;
> use Test;
> plan 2;
>
> # P28 (**) Sorting a list of lists according to length of sublists
> #
> # a) We suppose that a list contains elements that are lists
> themselves. The
> # objective is to sort the elements of this list according to their
> length. E.g.
> # short lists first, longer lists later, or vice versa.
> #
> # Example:
> # * (lsort '((a b c) (d e) (f g h) (d e) (i j k l) (m n) (o)))
> # ((O) (D E) (D E) (M N) (A B C) (F G H) (I J K L))
> #
> # b) Again, we suppose that a list contains elements that are lists
> themselves.
> # But this time the objective is to sort the elements of this list
> according to
> # their length frequency; i.e., in the default, where sorting is done
> # ascendingly, lists with rare lengths are placed first, others with
> a more
> # frequent length come later.
> #
> # Example:
> # * (lfsort '((a b c) (d e) (f g h) (d e) (i j k l) (m n) (o)))
> # ((i j k l) (o) (a b c) (f g h) (d e) (d e) (m n))
> #
> # Note that in the above example, the first two lists in the result
> have length 4
> # and 1, both lengths appear just once. The third and forth list have
> length 3
> # which appears twice (there are two list of this length). And
> finally, the last
> # three lists have length 2. This is the most frequent length.
> #
> # Arithmetic
>
>
> my @input= [<a b c>],[<d e>],[<f g h>],[<d e>],[<i j k l>],[<m
> n>],[<o>];
> my @expected= [<o>],[<d e>],[<d e>],[<m n>],[<a b c>],[<f g h>],[<i j
> k l>];
>
> # we could use
> # sort: {+$_}
> # but pugs seem to not support this yet
>
> my @[EMAIL PROTECTED]: {+$^a <=> +$^b};
> is @expected,
> @sorted,
> "We should be able to sort a list of lists according to length of
> sublists";
>
> # the list is not the same as in the sample text, when two lists have
> the
> # same frequency of length the ordering is unspecified, so this
> should be ok
>
> @expected= [<o>],[<i j k l>],[<a b c>],[<f g h>],[<d e>],[<d e>],[<m
> n>];
>
> # group lists by length
>
> my %grouped;
> for (@input) {%grouped{+$_}.push($_)}
>
> # now sort the values by frequency, again can't use
> # sort: {+$_}
>
> @sorted= %grouped.values.sort: {+$^a <=> +$^b};
> is @expected,@sorted, "..or according to frequency of length of
> sublists"
>
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