From: Robert Klemme <shortcutter@...>
Date: 2009-06-08T18:27:39+09:00
Subject: Re: How to sort array ascending, except zero ?

2009/6/8 Martin DeMello <martindemello@gmail.com>:
> On Mon, Jun 8, 2009 at 2:10 PM, Paganoni<noway@fakenullvoid.com> wrote:
>> Hello, I need to sort
>> [1,4,2,0,8,9] to [1,2,4,8,9,0]
>>
>> A simple ascending sort but with the zero values to the end
>
> max = a.max + 1
> # or max = 2 ** 31, say, if you don't want the extra pass
> # but sorting is O(n log n) and max is O(n) so it doesn't really matter

Still traversing the array twice just to get the max beforehand does
not /feel/ right.  I'd rather use your "large constant" - maybe even
with a _really_ large number: :-)

irb(main):020:0> a = [1,4,2,0,8,9]
=> [1, 4, 2, 0, 8, 9]
irb(main):021:0> INF = 1.0 / 0.0
=> Infinity
irb(main):022:0> a.sort_by {|x| x == 0 ? INF : x}
=> [1, 2, 4, 8, 9, 0]

Another good alternative is to use the block form of #sort:

irb(main):023:0> a.sort do |x,y|
irb(main):024:1*   case
irb(main):025:2*   when x == 0 then  1
irb(main):026:2>   when y == 0 then -1
irb(main):027:2>   else             x <=> y
irb(main):028:2>   end
irb(main):029:1> end
=> [1, 2, 4, 8, 9, 0]

Kind regards

robert


-- 
remember.guy do |as, often| as.you_can - without end
http://blog.rubybestpractices.com/