From: mike.leddy@...
Date: 2007-02-09T22:25:12+09:00
Subject: Re: Ruby hash equlity

Thanks Phrogz, but yes my problem would require nested hashes and
would have
to be reasonably efficient. Let me explain..... I started with the
problem of comparing
sets that may have sets as members:

irb(main):017:0> require 'set'
=> false
irb(main):018:0> Set[:a, :b] == Set[:a, :b]
=> true
irb(main):019:0> Set[:a, :b, Set[:c, :d]] == Set[:a, :b, Set[:c, :d]]
=> false

Suprisingly it didn't work. So I looked at how sets were implemented -
basically
hashes where the set members are the keys and the values are true.

I still believe that hash equality should be as the 'ri' manual
describes. I'm going
to examine the ruby 'C'  code to see exactly why.....

Mike



On Feb 8, 8:03 pm, "Phrogz" <g...@refinery.com> wrote:
> Finally, here's a slightly better version. It's still O(n^2) in the
> worst case, but should perform much better under common circumstances
> (in case someone actually needed this functionality):
>
> class Hash
>   # Ouch! O(n^2) performance
>   # Bad for hashes with many keys
>   def sort_of_equal( other )
>     equal = true
>     self.each{ |k,v|
>       unless found_key = (other[k]==v)
>         other.each{ |k2,v2|
>           break if found_key = ( ( k==k2 )  && ( v==v2 ) )
>         }
>       end
>       equal &&= found_key
>     }
>     equal
>   end
> end
>
> Note that this will still not work if you have nested hashes that you
> want to treat like this. You'd need to override Hash#== fully, or put
> in tests based on key and value type.