Forum: Ruby Noob: The ruby way to multiple Array by multiple?

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Eric L. (Guest)
on 2006-03-28 14:15
(Received via mailing list)
I want to make a Array multipled by another Array, and the following
semantic.
For example:
 [1, 2, 3] * [4, 5]
then I will want [[1, 4], [1, 5], [2, 4], [2, 5], [3, 4], [3, 5]]
returned.

I have the following code snippet to implement this function:

class Array
  alias_method :orig_multiple, :*
  def *(arg)
    case arg
    when Array
      if arg.empty?
        self
      else
        inject([]) do |product, item|
          product.concat(arg.collect do |i|
                           case item
                           when Array
                             item.clone << i
                           else
                             [item, i]
                           end
                         end)
        end
      end
    else
      orig_multiple(arg)
    end
  end
end

It works but seems ugly.

Could anyone please to give me a clue to make it beauty and rubish!

Thx in advance.
Eric
Robert K. (Guest)
on 2006-03-28 14:30
(Received via mailing list)
Eric L. wrote:
>     case arg
>                              [item, i]
> It works but seems ugly.
>
> Could anyone please to give me a clue to make it beauty and rubish!
>
> Thx in advance.
> Eric

res = []
a1.each {|e1| a2.each {|e2| res << [e1,e2]}}

Or, if you prefer inject

a1.inject([]) {|arr,e1| a2.each {|e2| arr << [e1,e2]}; arr }

Kind regards

	robert
Ryan D. (Guest)
on 2006-03-28 14:55
(Received via mailing list)
On Mar 28, 2006, at 2:08 AM, Eric L. wrote:

> I want to make a Array multipled by another Array, and the
> following semantic.
> For example:
>  [1, 2, 3] * [4, 5]
> then I will want [[1, 4], [1, 5], [2, 4], [2, 5], [3, 4], [3, 5]]
> returned.

a = [1, 2, 3]
b = [4, 5]

(a * b.size).zip(b * a.size).sort
# => [[1, 4], [1, 5], [2, 4], [2, 5], [3, 4], [3, 5]]

# or w/o sort:
a.zip(a).flatten.zip(b * a.size)
# => [[1, 4], [1, 5], [2, 4], [2, 5], [3, 4], [3, 5]]

# the latter only works as-is because b.size == 2 (a.zip(a).flatten
~= a * 2).
# If b.size were 3, you'd add b.size-1 a's to the first zip arg list:

b << 6
a.zip(a, a).flatten.zip(b * a.size)
# => [[1, 4], [1, 5], [1, 6], [2, 4], [2, 5], [2, 6], [3, 4], [3, 5],
[3, 6]]

# if order doesn't actually matter, the first solution is the cleanest:

class Array
   def x(b)
     a = self
     (a * b.size).zip(b * a.size)
   end
end

a.x b
=> [[1, 4], [2, 5], [3, 6], [1, 4], [2, 5], [3, 6], [1, 4], [2, 5],
[3, 6]]
Pit C. (Guest)
on 2006-03-28 16:32
(Received via mailing list)
Ryan D. schrieb:
>     a = self
>     (a * b.size).zip(b * a.size)
>   end
> end
>
> a.x b
> => [[1, 4], [2, 5], [3, 6], [1, 4], [2, 5], [3, 6], [1, 4], [2, 5],  [3,
> 6]]

Ryan, this works only if a.size and b.size are relatively prime, as in
your first example. The second example shows the error. If you use sort
in a different place it should work, though:

   (a * b.size).sort.zip(b * a.size)

Regards,
Pit
Eric L. (Guest)
on 2006-03-29 08:09
(Received via mailing list)
Ryan> class Array
  Ryan> def x(b)
  Ryan> a = self
  Ryan> (a * b.size).zip(b * a.size)
  Ryan> end
  Ryan> end

  Ryan> a.x b
  Ryan> => [[1, 4], [2, 5], [3, 6], [1, 4], [2, 5], [3, 6], [1, 4], [2,
5],
  Ryan> [3, 6]]

  Thanks for your hint, It seems so simple and pretty.
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