Subject: Re: [xsl] A superefficient way to compute the sum of A[i] * B[i] for i=1 to n? From: "Michael Kay mike@xxxxxxxxxxxx" <xsllistservice@xxxxxxxxxxxxxxxxxxxxxx> Date: Sat, 9 May 2020 13:55:28 0000 
I doubt you'll find much improvement on this. You could cut out the call on number() and rely on implicit conversion, but I doubt it makes any difference. You could factor out the expressions ($A/col) and ($B/col) into variables declared outside the loop, which might make a difference: finding the Nth child of an element might well take time proportional to N, whereas finding the Nth item in a sequence held in a variable is likely to be constant time. But it depends on the processor, of course. Measgre it and let us know the results. A significant part of the cost is likely to be stringtodouble conversion, and there's no way of avoiding that. Michael Kay Saxonica > On 9 May 2020, at 12:59, Costello, Roger L. costello@xxxxxxxxx <xsllistservice@xxxxxxxxxxxxxxxxxxxxxx> wrote: > > Hi Folks, > > I need a superefficient way to compute the sum of A[i] * B[i] for i=1 to n. > > For example, suppose A is this: > > <row> > <col>0.9</col> > <col>0.3</col> > </row> > > and B is this: > > <row> > <col>0.2</col> > <col>0.8</col> > </row> > > I want to compute: > > (0.9 * 0.2) + (0.3 * 0.8) > > Here's one way to do it: > > sum(for $i in 1 to count($A/col) return number($A/col[$i]) * number($B/col[$i])) > > I suspect that is not the most efficient approach. > > What is the most efficient approach? I will be doing hundreds of thousands of these computations, so I want to use the most efficient approach. > > /Roger
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