Forum: Ruby Multiplying array

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WKC C. (Guest)
on 2007-01-23 17:05
HI,

Does anyone know of a function that multiplies the contents of an array.
For example:

one = [1,2,3]
two = [[2],[3],[4]]

output = [[2],[6],[12]]

I've written a simple function that does this, however I'm sure there is
a better way, instead of casting the item to a float.

def multiplyArray(arr1,arr2)
  newArr = []
  i=0
  arr1.each do |x|
    x = x.to_s.to_f
    item = arr2[i].to_s.to_f
    ele = x * item
    newArr.push(ele)
    i = i + 1
   end
  puts newArr
end

Thanks,
Robert K. (Guest)
on 2007-01-23 17:25
(Received via mailing list)
On 23.01.2007 16:05, WKC CCC wrote:
> Does anyone know of a function that multiplies the contents of an array.
> For example:
>
> one = [1,2,3]
> two = [[2],[3],[4]]
>
> output = [[2],[6],[12]]
>
> I've written a simple function that does this, however I'm sure there is
> a better way, instead of casting the item to a float.

Why do you cast to float?  You can multiply integers directly.

First with a straightforward array "two":

irb(main):001:0> require 'enumerator'
=> true
irb(main):002:0> one = [1,2,3]
=> [1, 2, 3]
irb(main):003:0> two = [2,3,4]
=> [2, 3, 4]
irb(main):004:0> one.to_enum(:zip, two).map {|a,b| a*b}
=> [2, 6, 12]

Now with your array:

irb(main):005:0> two.map! {|i| [i]}
=> [[2], [3], [4]]
irb(main):006:0> one.to_enum(:zip, two).map {|a,b| a * b[0]}
=> [2, 6, 12]

Kind regards

  robert
unknown (Guest)
on 2007-01-23 17:36
(Received via mailing list)
On Wed, 24 Jan 2007, WKC CCC wrote:

> I've written a simple function that does this, however I'm sure there is
>    i = i + 1
>   end
>  puts newArr
> end
>
> Thanks,

if you are doing lots of this then check out narray:

   harp:~ > cat a.rb
   require 'narray'

   a = NArray.to_na [1,2,3]
   b = NArray.to_na [2,3,4]

   p( a * b )

   harp:~ > ruby a.rb
   NArray.int(3):
   [ 2, 6, 12 ]

regards.

-a
Gavin K. (Guest)
on 2007-01-23 18:26
(Received via mailing list)
WKC CCC wrote:
> Does anyone know of a function that multiplies the contents of an array.

module Enumerable
  def product
    inject{ |piece, prod| prod*piece }
  end
  def sum
    inject(0){ |piece, total| total+piece }
  end
end

a = (1..10)
p a.sum
#=> 55
p a.product
#=> 3628800
Rob B. (Guest)
on 2007-01-23 18:34
(Received via mailing list)
On Jan 23, 2007, at 11:25 AM, Phrogz wrote:

>   end
> end
>
> a = (1..10)
> p a.sum
> #=> 55
> p a.product
> #=> 3628800

module Enumerable
   def product
     inject(1) { |prod, piece| prod*piece }
   end
   def sum
     inject(0) { |total, piece| total+piece }
   end
end

True, except the block parameters were reversed.  In these cases, the
final result wasn't affected because the operations are commutative.

I also added the argument to the inject within product for symmetry.
You could equally remove the 0 argument from the sum method's use of
inject.

-Rob

Rob B.    http://agileconsultingllc.com
removed_email_address@domain.invalid
Gavin K. (Guest)
on 2007-01-23 18:51
(Received via mailing list)
Rob B. wrote:
> final result wasn't affected because the operations are commutative.
Man, I swear that every time I use #inject I get it backwards, and
think that I've fixed the answer in my mind for the next time. Thanks
for the correction. (Anyone got a good _why-like mnemonic they use for
remembering the order?)

> I also added the argument to the inject within product for symmetry.
> You could equally remove the 0 argument from the sum method's use of
> inject.

I thought about that, but in my mind, the sum of an empty array is
zero, but the product of an empty array is nil (or 0?), not 1. Of
course, totally up the the OP as to how s/he wants to handle this edge
case.
WKC C. (Guest)
on 2007-01-23 18:56
How can the inject function work on 2 arrays?
Patrick G. (Guest)
on 2007-01-23 18:57
[inject confusion]

 (Anyone got a good _why-like mnemonic they use for
> remembering the order?)

not good, but it helps me:

"memo" comes first (alphabetically), so it is

inject(memo, obj)


P.
Gavin K. (Guest)
on 2007-01-23 19:15
(Received via mailing list)
WKC CCC wrote:
> How can the inject function work on 2 arrays?

It can't; I only read the first line of your post (which described a
single array) and rushed off my answer. As penance, I give you another
solution:

a = [ 1, 2, [3], [4], 5]
b = [ 6, [7], [8], 9, 10 ]

p a.flatten.zip(b.flatten).map{ |a,b| a*b }
#=> [6, 14, 24, 36, 50]
Bertram S. (Guest)
on 2007-01-23 22:09
(Received via mailing list)
Hi,

Am Mittwoch, 24. Jan 2007, 01:57:36 +0900 schrieb Patrick G.:
> [inject confusion]
>
> > (Anyone got a good _why-like mnemonic they use for
> > remembering the order?)
>
> not good, but it helps me:
>
> "memo" comes first (alphabetically), so it is
>
> inject(memo, obj)

I'm trying to help myself with the order invariance in

  a == a.inject([]) { |r,e| r+[e] }
  a == a.inject([]) { |r,e| r << e }

Bertram
William J. (Guest)
on 2007-01-23 23:43
(Received via mailing list)
WKC CCC wrote:
> I've written a simple function that does this, however I'm sure there is
>     i = i + 1
>    end
>   puts newArr
> end
>
> Thanks,
>
> --
> Posted via http://www.ruby-forum.com/.

[1,2,3].zip([2,3,4]).map{|a,b| a*b}
    ==>[2, 6, 12]
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