From: "Guy N. Hurst" <gnhurst@...>
Date: 2002-03-19T20:30:56+09:00
Subject: Re: How to implement each_continuous_range

andrew queisser wrote:
> 
> I'd like to implement an iterator that returns all ranges of
> successive elements of an array. If an array consists of [1,2,3,5,6]
> the iterator should return 1,3 and 5,6. I don't care whether the
> iterator returns an array with two elements or two values. So far I
> have this:
> 
> #----------------------------------------
> x =  [ 1, 2, 3, 4, 7, 8, 12, 15, 16, 22 ]
> 
> class Array
>   def each_cont_range
>     r0 = 0
>     for i in 0..length()-2
>       if (self[i+1] != self[i].succ)
>         yield(self[r0], self[i])
>         r0 = i+1
>       end
>     end
>     yield(self[r0], self[-1])
>   end
> end
> 
> x.each_cont_range { |r0, r1| print "range #{r0} to #{r1}\n" }
> #---------------------------------------------------
> 
> It works but it isn't pretty. I tried to use an inject function but
> didn't get anywhere. Suggestions for improvement? Are there any
> built-in ways of doing this?
> 

I don't think there is any built-in way to do this, but here
is my contribution (I call it each_range):


class Array
  def each_range
    a=self.dup
    a.each_index{|i| yield(a[i],(a.slice!(i) while a[i+1]==a[i].succ;a[i]))}
  end
end


irb preview:
===========
irb(main):001:0> class Array
irb(main):002:1>   def each_range
irb(main):003:2>     a=self.dup
irb(main):004:2>     a.each_index{|i| yield(a[i],(a.slice!(i) while a[i+1]==a[i].succ;a[i]))}
irb(main):005:2>   end
irb(main):006:1> end
nil
irb(main):007:0> x =  [ 1, 2, 3, 4, 7, 8, 12, 15, 16, 22 ]
[1, 2, 3, 4, 7, 8, 12, 15, 16, 22]
irb(main):008:0> x.each_range { |r0, r1| puts "range #{r0} to #{r1}" }
range 1 to 4
range 7 to 8
range 12 to 12
range 15 to 16
range 22 to 22
[4, 8, 12, 16, 22]
irb(main):009:0>


I made it return arrays of start,end pairs because you used it
that way, but it is a simple matter to return a Range object, too.


Guy N. Hurst