Re: Is high-speed sorting impossible with Ruby?

On 29.11.2011 02:05, Josh Cheek wrote:
> 2011/11/28 Matthias Wächter<[e-mail]>
>
>> A similar solution, at first glance only marginally different, is the
>> following:
>>
>>   $stdin.gets
>>> puts $stdin.each_line.sort_by(&:to_**i)
>>>
>>
>> Besides the fact that this solution is slower than yours, it uses a neat
>> little technique of Ruby 1.9, Enumerators.
>
>
> $stdin is already enumerable, so each_line isn't necessary
>
> puts $stdin.drop(1).sort_by(&:to_i)

Great one-liner, thanks!

But in the context of my posting, it is not following the Enumerator 
principle. $stdin.drop(1) creates a large array on its output which I 
wanted to avoid.

What we need is an enumeratoresque version of Enumerable#drop, which I 
call drop_first. It will only consume on its input

> class Enumerator
>   def filter1(&blk)
>     self.class.new do |y|
>       each do |*input|
>         y << blk.call(*input)
>       end
>     end
>   end
>
>   def drop_first(n)
>     self.class.new do |y|
>       with_index do |input,idx|
>         y << input if idx >= n
>       end
>     end
>   end
> end
>
> puts $stdin.to_enum(:each_line).drop_first(1).filter1(&:to_i).sort

Performance-wise not better at all, but a nice excursion.

– Matthias