Normalizing A Vector
in [Python]
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From: Alain Ketterlin on 2 Aug 2010 06:19 "Bartc" <bartc(a)freeuk.com> writes: >> def norm(V): >> L = math.sqrt( sum( [x**2 for x in V] ) ) >> return [ x/L for x in V ] > > There's a cost involved in using those fancy constructions. Sure. The above has three loops that take some time. > I found the following to be about twice as fast, when vectors are > known to have 3 elements: > > def norm3d(v): > L = math.sqrt((v[0]*v[0]+v[1]*v[1]+v[2]*v[2])) > return (v[0]/L,v[1]/L,v[2]/L) > > (Strangely, changing those divides to multiplies made it slower.) You mean by setting L to 1.0 / math.sqrt(...) and using v[0]*L etc.? I think * and / have the same cost on floats, and the added / adds some cost. But what you observe is probably caused by the overloading of "*", that needs more type checks. You may try with operator.mul to see if the call compensates the cost of type checking, but I doubt it. -- Alain.
From: Bartc on 2 Aug 2010 07:32 "Alain Ketterlin" <alain(a)dpt-info.u-strasbg.fr> wrote in message news:87fwyxgvuv.fsf(a)dpt-info.u-strasbg.fr... > "Bartc" <bartc(a)freeuk.com> writes: >> def norm3d(v): >> L = math.sqrt((v[0]*v[0]+v[1]*v[1]+v[2]*v[2])) >> return (v[0]/L,v[1]/L,v[2]/L) >> >> (Strangely, changing those divides to multiplies made it slower.) > > You mean by setting L to 1.0 / math.sqrt(...) and using v[0]*L etc.? Yes. > I think * and / have the same cost on floats, and the added / adds > some cost. I expected no measurable difference, not running Python anyway (I tried it in gcc and using divides increased runtimes by 50%, corresponding to some 1% for Python). I would naturally have written it using multiplies, and was just surprised at a 3-4% slowdown. > But what you observe is probably caused by the overloading of > "*", that needs more type checks. That sounds reasonable. -- Bartc
From: Lawrence D'Oliveiro on 2 Aug 2010 21:57 In message <6Dw5o.72330$Ds3.63060(a)hurricane>, Bartc wrote: > There's a cost involved in using those fancy constructions. Sure. But at the point that starts to matter, you have to ask yourself why you're not rewriting the CPU-intensive part in C.
From: sturlamolden on 2 Aug 2010 22:37
On 30 Jul, 13:46, Lawrence D'Oliveiro <l...(a)geek- central.gen.new_zealand> wrote: > Say a vector V is a tuple of 3 numbers, not all zero. You want to normalize > it (scale all components by the same factor) so its magnitude is 1. > > The usual way is something like this: > > L = math.sqrt(V[0] * V[0] + V[1] * V[1] + V[2] * V[2]) > V = (V[0] / L, V[1] / L, V[2] / L) > > What I dont like is having that intermediate variable L leftover after the > computation. L = math.sqrt(V[0] * V[0] + V[1] * V[1] + V[2] * V[2]) V = (V[0] / L, V[1] / L, V[2] / L) del L But this is the kind of programming tasks where NumPy is nice: V[:] = V / np.sqrt((V**2).sum()) Sturla |