Subject: Re: [xsl] A superefficient way to compute the sum of A[i] * B[i]for i=1 to n? From: "Costello, Roger L. costello@xxxxxxxxx" <xsllistservice@xxxxxxxxxxxxxxxxxxxxxx> Date: Sat, 9 May 2020 16:45:26 0000 
Thank you Michael, Michael, and Martin. I measured the performance of this: sum(for $i in 1 to count($A/col) return number($A/col[$i]) * number($B/col[$i])) and this: sum(for $i in 1 to count($A) return number($A [$i]) * number($B [$i])) in the latter, $A and $B holds the sequence of values in the <col> elements. I ran the two versions 16.6 million times. The first version (which involves finding the Nth child element) took: 0.670 seconds The second version (which involves finding the Nth item in a sequence) took: 0.852 seconds It is faster to find the Nth child element than to find the Nth item in a sequence  surprising! I used SAXON EE 9.1.4 /Roger From: Michael Kay mike@xxxxxxxxxxxx <xsllistservice@xxxxxxxxxxxxxxxxxxxxxx> Sent: Saturday, May 9, 2020 9:55 AM To: xsllist@xxxxxxxxxxxxxxxxxxxxxx Subject: [EXT] Re: [xsl] A superefficient way to compute the sum of A[i] * B[i]for i=1 to n? I doubt you'll find much improvement on this. You could cut out the call on number() and rely on implicit conversion, but I doubt it makes any difference. You could factor out the expressions ($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. mailto:costello@xxxxxxxxx <mailto: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 http://www.mulberrytech.com/xsl/xsllist http://lists.mulberrytech.com/unsub/xsllist/673357 ()
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