[lnkForumImage]
TotalShareware - Download Free Software
Confronta i prezzi di migliaia di prodotti.
Asp Forum
 Home | Login | Register | Search 


 

Forums >

comp.lang.ruby

Re: how to remove dups from 2 lists?

Harry Kakueki

5/11/2007 2:41:00 PM

On 5/11/07, Mike Steiner <mikejaysteiner@gmail.com> wrote:
> I'm trying to write some code that removes all elements from 2 lists that
> are in both lists. However, I don't want any duplicates from each list
> deleted also (which is what the array "-" operator does). The code I have
> now doesn't handle restarting the current iteration for both loops when a
> match is found and deleted in both loops. Here's the code:
>
> def RemoveDupsFromLists ( list1 , list2 )
> list1.each_index do | i |
> list2.each_index do | j |
> if list1[i] == list2[j]
> list1.delete_at ( i )
> list2.delete_at ( j )
> end
> end
> end
> return [ list1 , list2 ]
> end
>
> What's weird is that doing this is easy in C (my first language), but
> difficult in Ruby. Everything else I've seen has been MUCH easier in Ruby.
>
> Mike Steiner
>

After thinking about your question again, I think you meant something
a little different than what I was thinking before, I think :) :)
This is no shorter than your code, just different.

arr1 = %w[a b car car b c r c car c c r d]
arr2 = %w[a r1 a b c c car d r r r d]

counts1 = Hash.new(0)
arr1.each {|x| counts1[x] += 1}

counts2 = Hash.new(0)
arr2.each {|x| counts2[x] += 1}

new1 = []
new2 = []

arr1.uniq.each do |x|
(counts1[x] - counts2[x]).times {new1 << x} if counts1[x] > counts2[x]
end

arr2.uniq.each do |x|
(counts2[x] - counts1[x]).times {new2 << x} if counts2[x] > counts1[x]
end

p new1 #["b","car","car", "c", "c"]
p new2 #["a","r1","d","r"]

Harry

--
http://www.kakueki.com/ruby...
A Look into Japanese Ruby List in English