From: Mark Hubbart <discordantus@...>
Date: 2004-11-10T03:28:33+09:00
Subject: Re: Iterating trough hash

On Wed, 10 Nov 2004 02:03:45 +0900, ara.t.howard@noaa.gov
<ara.t.howard@noaa.gov> wrote:
> 
> 
> On Tue, 9 Nov 2004, Kevin [ISO-8859-15] B�rgens wrote:
> 
> > Hi!
> >> #!/usr/bin/env ruby
> >>
> >> h = {"john" => 41, "mary" => 31, "fred" => 10}
> >> h.each_key do |person1|
> >> h.each_key do |person2|
> >> puts "#{person1} shakes hands with #{person2}.  " +
> >> "Together they are #{h[person1] + h[person2]} years old."
> >> end
> >> end
> > No, this programm would iterate like this
> > john john
> > john mary
> > john fred
> > mary john
> > mary mary
> > mary fred
> > fred john
> > fred mary
> > fred fred
> >
> >> I'm not very clear on what you were aiming for with the if statement,
> >> but if you tell me, I can probably add that back a little slimmer too.
> > Therefore I used the example with shaking hands. You don't shake hands with
> > yourself and it doesn't matter who is person1 or person2
> >
> > TIA,
> > Kevin
> 
> that makes sense - but why, in your code, are you checking that the person's
> name AND/OR age are less than the other person?  eg this line
> 
>    if h.sort.index([person1,h[person1]]) < h.sort.index([person2,h[person2]])
> 
> says:
> 
>    if the name of person1 is less than the name of person2 OR the names are the
>    SAME but the age of person1 is less than the age of person2.

Um, no, I think this says:

  if the index of the first element that matches [person1, age1] is
greater than the
  index of the first element that matches [person2, age2].

> 
> example:
> 
>     harp:~ > irb
> 
>    irb(main):001:0> list = ['john', 42], ['john', 41]
>    => [["john", 42], ["john", 41]]
> 
>    irb(main):002:0> a, b = list.first, list.last
>    => [["john", 42], ["john", 41]]
> 
>    irb(main):003:0> list.index(a)
>    => 0
> 
>    irb(main):004:0> list.index(b)
>    => 1
> 
>    irb(main):005:0> list.index(a) < list.index(b)
>    => true
> 
> and john is shaking hands with john.  ruby sorts array by using this algorithim:
> 
>    compare the first elements, if tied
>      compare the second elements, if tied
>        compare the third elelments, etc....
>          etc....
> 
> so the comparison you are making by checking the sort order (index position of
> entry) checks both name AND age - which may or may not be what you want.

But, since this array starts out its life as a hash, there will never
be duplicate names. So the ages will never be compared.

cheers,
Mark

> 
> cheers and welcome to ruby!
> 
> 
> 
> -a
> --
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