Re: Improving performance of hash math

On Jan 19, 2011, at 6:13 PM, Ryan Davis wrote:
>=20
> I don't agree with Gary's suggestion, because in this case it is a =
proper application of inject, but if you think it is 2x slower, then =
you're benchmarking is probably at fault.

I'm OK with being wrong. I didn't actually try my own suggestion so I'm =
not going to defend it too hard but I did notice that your =
euclidian_distance2 version that uses #each was 30% faster than =
euclidian_distance1 that used #inject.  Your euclidian_distance3 version =
uses both #inject and #each but would seem to be sensitive to the size =
of the lhs argument (i.e. the number of entries in the lhs hash).

I'd like to see the sample data to better understand what is going on.

Gary Wright


>=20
> My measurements show each to be roughly equivalent on 19 and quite a =
bit faster on 1.8. That could be explained by my modification to your =
treatment of rhs or it could also be explained by my fake data.
>=20
> % ./q.rb 10000
> # of iterations =3D 10000
>                          user     system      total        real
> null_time             0.000000   0.000000   0.000000 (  0.001118)
> euclidian_distance1  18.960000   0.010000  18.970000 ( 19.011762)
> euclidian_distance2  13.050000   0.010000  13.060000 ( 13.069581)
> euclidian_distance3  13.090000   0.010000  13.100000 ( 13.109778)
>=20
> against:
>=20
> def euclidian_distance1(lhs, rhs)
>   @s =3D lhs.inject(0.0) do |s, (k,v)|
>     rhs.has_key?(k) ? (s + (v - rhs[k])**2) : (s + v**2)
>   end
>   @s =3D rhs.inject(@s) do |s, (k,v)|
>     lhs.has_key?(k) ? s : (s + v**2)
>   end
>   @s
> end
>=20
> def euclidian_distance2(lhs, rhs)
>  s =3D 0.0
>  lhs.each do |k,v|
>    s +=3D rhs.has_key?(k) ? ((v - rhs[k])**2) : (v**2)
>  end
>  rhs.each do |k,v|
>    s +=3D v**2 if lhs.has_key?(k)
>  end
>  s
> end
>=20
> def euclidian_distance3(lhs, rhs)
>  s =3D lhs.inject(0.0) do |n, (k,v)|
>    n + (rhs.has_key?(k) ? v - rhs[k] : v)**2
>  end
>  rhs.each do |k,v|
>    s +=3D v**2 if lhs.has_key? k
>  end
>  s
> end
>=20
>=20