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" > -- Buy the book -- http://www.oreilly.com/catalog/perlhks/ Perl and CGI -- http://users.easystreet.com/ovid/cgi_course/