       Re: Speed of dot product in Mathematica

• To: mathgroup at smc.vnet.net
• Subject: [mg5277] Re: [mg5230] Speed of dot product in Mathematica
• From: BobHanlon at aol.com
• Date: Mon, 18 Nov 1996 02:30:24 -0500
• Sender: owner-wri-mathgroup at wolfram.com

```You can also use

Plus @@ (a b)

If you want the dot product of portions of the factors, use only the
portions.  For example,

Take[a, {100, 1000}].Take[b, {200, 1100}]

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Subj:  [mg5230] Speed of dot product in Mathematica
To: mathgroup at smc.vnet.net
X-From: carlos at mars.Colorado.EDU (Carlos A. Felippa)
To: mathgroup at smc.vnet.net

The speed of dot products in Mathematica 2.2 depends
significantly on implementation.  For example, timing

n=10000; a=Table[1.,{n}]; b=a;
Print[Timing[a.b]];
Print[Timing[s=Sum[a[[i]]*b[[i]],{i,1,n}]]]
Print[Timing[s=0;Do[s+=a[[i]]*b[[i]],{i,1,n}]]];
Print[Timing[s=0;For[i=0,i<=n,i++,s+=a[[i]]*b[[i]] ]]];

on a Mac 8500 gives

{0.0666667 Second, 10000.}
{1.48333 Second, 10000.}
{2.38333 Second, Null}
{3.95 Second, Null}

Is there a way to speed up the Sum form, for example using Compile, so that
it achieves a performance similar to that of the built-in dot operator?

This is important in matrix routines where the dot product involves only
portions of rows or columns, or where the stride is not unity.

BTW, several of the LinearAlgebra Package functions (e.g. lufactor)
use the For-loop implementation.  As shown above, that has the worst
performance, being 60 times slower than the built-in operator.

```

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