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iterations in a hash

by madM (Beadle)
on Aug 04, 2013 at 14:36 UTC ( #1047798=perlquestion: print w/ replies, xml ) Need Help??
madM has asked for the wisdom of the Perl Monks concerning the following question:

Hi! Iīm trying to compute an iteration process with a hash for example if i have this hash:
%hash=( A=>4, B=>5, C=>10, D=>2, E=>9 )
i want to multiply the A with the B , then the A with the C .. A with D .. A with E.. and then B with A and so on.. i have been trying to compute that without success .. any help would be really appreciated Thanks!

Comment on iterations in a hash
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Re: iterations in a hash
by rjt (Deacon) on Aug 04, 2013 at 14:54 UTC

    Hello madM! Welcome to PerlMonks.

    i want to multiply the A with the B , then the A with the C .. A with D .. A with E.. and then B with A and so on..

    I'm really not sure what you mean. What output do you expect? The only non-trivial (i.e., pure repetition) case I can think of is that you want to see each product of A*B, A*C, A*D, …, printed separately, like this:

    A * B = 20 A * C = 40 A * D = 8 A * E = 36 B * A = 20 B * C = 50 : :

    Is that what you want? If so, I suggest you have a look at keys, printf, and Foreach Loops, and ponder the following:

    for my $x (sort keys %hash) { for my $y (sort grep { $_ ne $x } keys %hash) { printf "%s * %s = %d\n", $x, $y, $hash{$x} * $hash{$y}; } }

    If you are looking for a different result, please give us an example of the output you are expecting.

Re: iterations in a hash
by BrowserUk (Pope) on Aug 04, 2013 at 15:06 UTC

    That is an inherently recursive process:

    sub crossMul{ my( $r, $first ) = ( shift, shift ); print "$first * $_ = ", $r->{$first} * $r->{$_} for @_; crossMul( $r, @_ ) if @_ > 1; };; %hash=( A=>4, B=>5, C=>10, D=>2, E=>9 );; crossMul( \%hash, sort keys %hash );; A * B = 20 A * C = 40 A * D = 8 A * E = 36 B * C = 50 B * D = 10 B * E = 45 C * D = 20 C * E = 90 D * E = 18

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Re: iterations in a hash
by Loops (Hermit) on Aug 04, 2013 at 15:10 UTC
    my %hash=(A=>4, B=>5, C=>10, D=>2, E=>9); my @zip = map { [ $_, $hash{$_} ] } sort keys %hash; do { my $f = shift @zip; print ("@$f * @$_ = ", $f->[1] * $_->[1], "\n") for (@zip); } while (@zip);
    Or alternatively being a smidgen easier to read:
    my %hash=(A=>4, B=>5, C=>10, D=>2, E=>9); my @zip = sort keys %hash; do { my $f = shift @zip; for (@zip) { my ($v1,$v2) = @hash{$f,$_}; print "$f($v1) * $_($v2) = ", $v1 * $v2, "\n"; } } while (@zip);
    A(4) * B(5) = 20 A(4) * C(10) = 40 A(4) * D(2) = 8 A(4) * E(9) = 36 B(5) * C(10) = 50 B(5) * D(2) = 10 B(5) * E(9) = 45 C(10) * D(2) = 20 C(10) * E(9) = 90 D(2) * E(9) = 18
Re: iterations in a hash
by Laurent_R (Vicar) on Aug 04, 2013 at 19:12 UTC

    i want to multiply the A with the B , then the A with the C .. A with D .. A with E.. and then B with A and so on.. i have been trying to compute that without success

    If you do that, you'll end up with all values of the hash being the product 4*5*10*2*9. I doubt that's what you want. Or, probably I did not really understand what you exactly want. Or perhaps you want to end up with a two dimentional table? Please explain.

Re: iterations in a hash
by kcott (Abbot) on Aug 05, 2013 at 09:10 UTC

    G'day madM,

    Welcome to the monastery.

    As others have pointed out, this is a poorly worded question: you tell from all the guesswork and requests for more information. Better questions get better answers: "How do I post a question effectively?" explains how to achieve this.

    Here's my take on a solution:

    $ perl -Mstrict -Mwarnings -E ' my %hash = (A => 4, B => 5, C => 10, D => 2, E => 9); local $" = "\t"; for my $key (sort keys %hash) { say "@{[map { qq{$key * $_ = } . $hash{$key} * $hash{$_} } sort grep { $_ ne $key } keys %hash]}" } ' A * B = 20 A * C = 40 A * D = 8 A * E = 36 B * A = 20 B * C = 50 B * D = 10 B * E = 45 C * A = 40 C * B = 50 C * D = 20 C * E = 90 D * A = 8 D * B = 10 D * C = 20 D * E = 18 E * A = 36 E * B = 45 E * C = 90 E * D = 18

    -- Ken

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