In Perl, how can I iterate over the Cartesian product of multiple sets?

本秂侑毒 提交于 2019-11-29 14:53:07

My Set::CrossProduct module does exactly what you want. Note that you aren't really looking for permutations, which is the ordering of the elements in a set. You're looking for the cross product, which is the combinations of elements from different sets.

My module gives you an iterator, so you don't create it all in memory. You create a new tuple only when you need it.

use Set::Crossproduct;

my $iterator = Set::CrossProduct->new(
    [
        [qw( foo bar baz )],
        [qw( cat dog     )],
        [qw( 1 2 3 4     )],
    ]
    );

while( my $tuple = $iterator->get ) {
    say join ' ', $tuple->@*;
    }

A simple recursive solution for an arbitrary number of lists:

sub permute {
  my ($first_list, @remain) = @_;

  unless (defined($first_list)) {
    return []; # only possibility is the null set
  }

  my @accum;
  for my $elem (@$first_list) {
    push @accum, (map { [$elem, @$_] } permute(@remain));
  }

  return @accum;
}

A not-so-simple non-recursive solution for an arbitrary number of lists:

sub make_generator {
  my @lists = reverse @_;

  my @state = map { 0 } @lists;

  return sub {
    my $i = 0;

    return undef unless defined $state[0];

    while ($i < @lists) {
      $state[$i]++;
      last if $state[$i] < scalar @{$lists[$i]};
      $state[$i] = 0;
      $i++;
    }

    if ($i >= @state) {
      ## Sabotage things so we don't produce any more values
      $state[0] = undef;
      return undef;
    }

    my @out;
    for (0..$#state) {
      push @out, $lists[$_][$state[$_]];
    }

    return [reverse @out];
  };
}

my $gen = make_generator([qw/foo bar baz/], [qw/cat dog/], [1..4]);
while ($_ = $gen->()) {
  print join(", ", @$_), "\n";
}

Recursive and more-fluent Perl examples (with commentary and documentation) for doing the Cartesian product can be found at http://www.perlmonks.org/?node_id=7366

Example:

sub cartesian {
    my @C = map { [ $_ ] } @{ shift @_ };

    foreach (@_) {
        my @A = @$_;

        @C = map { my $n = $_; map { [ $n, @$_ ] } @C } @A;
    }

    return @C;
}

There's one method I thought of first that uses a couple for loops and no recursion.

  1. find total number of permutations
  2. loop from 0 to total_permutations-1
  3. observe that, by taking the loop index modulus the number of elements in an array, you can get every permutations

Example:

Given A[3], B[2], C[3],

for (index = 0..totalpermutations) {
    print A[index % 3];
    print B[(index / 3) % 2];
    print C[(index / 6) % 3];
}

where of course a for loop can be substituted to loop over [A B C ...], and a small part can be memoized. Of course, recursion is neater, but this might be useful for languages in which recursion is severely limited by stack size.

You can use nested loops.

for my $e1 (qw( foo bar baz )) {
for my $e2 (qw( cat dog )) {
for my $e3 (qw( 1 2 3 4 )) {
   my @choice = ($e1, $e2, $e3); 
   ...
}}}

When you need an arbitrary number of nested loops, you can use Algorithm::Loops's NestedLoops.

use Algorithm::Loops qw( NestedLoops );

my @lists = (
   [qw( foo bar baz )],
   [qw( cat dog )],
   [qw( 1 2 3 4 )],
);

my $iter = NestedLoops(\@lists);
while ( my @choice = $iter->() ) {
   ...
}
易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!