Re: [xsl] A super-efficient way to compute the sum of A[i] * B[i] for i=1 to n?

["Martin Honnen martin.honnen@xxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>]

Am 09.05.2020 um 13:59 schrieb Costello, Roger L. costello@xxxxxxxxx:
> Hi Folks,
>
> I need a super-efficient 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.

You can express it a bit more nicely in XPath 3 with higher-order
functions as

  sum(for-each-pair($A/col, $B/col, function($a, $b) { $a * $b}))

but whether that performs better or worse is something you can only measure.