On Wed, Sep 2, 2009 at 1:41 AM, Max Williams<toastkid.williams / gmail.com> wrote:
>
>
> I have a situation where i have an array of 12 items.  If someone
> chooses to have n of them (where n can be between 3 and 12) then i want
> to always include the first and last, and then 'spread' the others out
> as evenly as possible between the rest.
>
> So, lets say for the sake of argument that the array holds the numbers 1
> to 12.
>
>>> arr = (1..12).to_a
> => [1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]
>
> I would get results back like this
>
> arr.spread(3)
> => [1,6,12] (or [1,7,12], either is fine)
>
> arr.spread(4)
> => [1, 5, 9, 12]  (or [1,4,8,12] or [1, 5, 8, 12])
>
>It feels like there should be a simple solution for this but i can't
>think of a nice way.  Anyone?
>
>thanks
>max



Thanks to a post by David Masover in a recent thread, I learned about
Object#tap.
He used it on a hash.

I've used it here on an array.
Is there any problem using Object#tap in this way?
I saw some examples, but in the examples the object was not modified
in the block.



class Array
 def spread(n)
   dup.tap{|a| (size-n).downto(1).map{|b| size*b/(size-n+1)}.each{|c|
a.delete_at(c)}}
 end
end

arr = (1..12).to_a
(3..12).each{|t| p arr.spread(t)}



###Output

[1, 6, 12]
[1, 4, 8, 12]
[1, 3, 6, 9, 12]
[1, 3, 5, 8, 10, 12]
[1, 2, 4, 6, 8, 10, 12]
[1, 2, 4, 6, 7, 9, 11, 12]
[1, 2, 3, 5, 6, 8, 9, 11, 12]
[1, 2, 3, 4, 6, 7, 8, 10, 11, 12]
[1, 2, 3, 4, 5, 6, 8, 9, 10, 11, 12]
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12]


Harry