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simultaneously adding and multiplying numbers in a hash/ array

by derpp (Acolyte)
on Aug 09, 2010 at 21:13 UTC ( #853917=perlquestion: print w/ replies, xml ) Need Help??
derpp has asked for the wisdom of the Perl Monks concerning the following question:

Hi guys,

I'm kind of stuck on a stupid question. I need to multiply a number by another number, then add it to the previous number..numerous times.

Here's a made up version of my code, since I'm extracting the data I'm using from MS excel and it would take way too long to type all that up.

use warnings; my @change = (1.15, -0.1, 5.4, 1.03, -0.241); my @numberofshares = (100, 400, 200, 300, 240); @hash{@change} = @numberofshares; while (($k, $v) = each %hash) { foreach ($v1) { $product = $v1*$k1 } }

Yes, I know this gives the incorrect number, but what I want to do is after each $k*$v, i'd add it to the previous result of $k*$v, and on and on.

I'm only a beginner at this, so please excuse my incorrect spacing or whatever in my code. Thanks in advance!

Comment on simultaneously adding and multiplying numbers in a hash/ array
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Re: simultaneously adding and multiplying numbers in a hash/ array
by toolic (Chancellor) on Aug 09, 2010 at 21:20 UTC
      Thanks! I can't believe the answer was so simple..
Re: simultaneously adding and multiplying numbers in a hash/ array
by psini (Deacon) on Aug 09, 2010 at 21:22 UTC

    Not sure I've understand the problem. You want the sum of the products of a set of pairs, right?

    If so, you don't need a hash, you could do:

    use strict; use warnings; my @change = (1.15, -0.1, 5.4, 1.03, -0.241); my @numberofshares = (100, 400, 200, 300, 240); my $sum=0; while (my $k=pop @change) { $sum+=$k*pop @numberofshares; } print $sum;

    Warning: untested

    Rule One: "Do not act incautiously when confronting a little bald wrinkly smiling man."

Re: simultaneously adding and multiplying numbers in a hash/ array
by ikegami (Pope) on Aug 09, 2010 at 21:44 UTC
    The hash is neither required nor helpful.
    my @change = (1.15, -0.1, 5.4, 1.03, -0.241); my @numberofshares = (100, 400, 200, 300, 240); my $product = 0; for (0..$#change) { $product += $change[$_] * $numberofshares[$_]; }
Re: simultaneously adding and multiplying numbers in a hash/ array
by AnomalousMonk (Abbot) on Aug 09, 2010 at 22:12 UTC

    The use of a hash in this application is positively dangerous unless the elements of the  @change array (which become the keys of the hash) can be guaranteed to be unique. E.g.:

    >perl -wMstrict -le "my @change = (1.15, -0.1, 5.4, 1.03, 1.15, -0.241); my @numberofshares = (9999, 400, 200, 300, 500, 240); my %hash; @hash{@change} = @numberofshares; printf qq{%d keys in hash (oops...) \n}, scalar keys %hash; my $sum = 0; while (my($k, $v) = each %hash) { $sum += $v * $k; } print qq{sum == $sum (?)}; " 5 keys in hash (oops...) sum == 1866.16 (?)

    Update: The algorithm can be expressed more concisely with some list-processing functions:

    >perl -wMstrict -le "use List::Util qw(sum); use List::MoreUtils qw(pairwise); use vars qw($a $b); my @change = (1.15, -0.1, 5.4, 1.03, -0.241); my @numberofshares = ( 100, 400, 200, 300, 240); my $sum = sum pairwise { $a * $b } @change, @numberofshares; print qq{sum == $sum}; " sum == 1406.16

    I think it could be expressed yet more concisely in Perl 6, but my 6-fu is not that great. Would it be something like:
        my $sum +=<<*<< @change, @numberofshares;

      >> I think it could be expressed yet more concisely in Perl 6

      I like:

      say [+] @change <<*>> @numberofshares
Re: simultaneously adding and multiplying numbers in a hash/ array
by JavaFan (Canon) on Aug 10, 2010 at 12:18 UTC
    my $sum = 0; for (my $i = 0; $i < @change; $i++) { $sum += $change[$i] * $numberofshares[$i]; }

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