From: "pauldacus@..." <pauldacus@...>
Date: 2005-10-18T02:41:56+09:00
Subject: Re: How to get non-unique elements from an array?

Well, until someone catches the horrible gaff on my last try by solving
the EXACT OPPOSITE PROBLEM WHILE BEING CONDESCENDING (my wife is used
to this), I'll keep trying:

Of course, the following is what works, finding duplicates ELEMENTS:

a.sort.to_s.scan(/(.)(\1)+/).flatten.uniq

Yeah... I guess it's clear this finds individual duplicate ELEMENTS,
not numbers.  Works on single digits/letters/etc, blows up on all else.
 How's that for useless?

What would do the trick is a 'non-greedy' array subtraction:

Normal(greedy): [0,1,2,3,4,5,2,3] - [0,1,2,3,4,5] = []
Non-greedy: [0,1,2,3,4,5,2,3] - [0,1,2,3,4,5] = [2,3]

Seems like there has got to be a way of solving this w/o counters or
blocks...

Sam Kong wrote:
> Hello!
>
> I need to get non-unique elements from an array.
> The best I came up with was using a hash as a counter for each unique
> elements.
>
> a = [0,1,2,3,4,5,2,3]
>
> #What I want to get is [2,3] as 2,3 are non-unique elements.
>
> h = {}
> a.each do |i|
>   if h[i]
>     h[i] += 1
>   else
>     h[i] = 1
>   end
> end
>
> u = []
> h.each do |k, v|
>   if v > 1
>     u << k
>   end
> end
>
> #now u == [2,3]
>
> This works fine.
> But I think there's a better way.
> How do you handle such a case?
> 
> Thanks in advance.
> 
> Sam