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 $B holds the sequence of values in the <col> elements.
I ran the two versions 16.6 million times.
The first version (which involves finding the Nth child element) took: 0.670
seconds
The second version (which involves finding the Nth item in a sequence) took:
0.852 seconds
It is faster to find the Nth child element than to find the Nth item in a
sequence - surprising!
I used SAXON EE 9.1.4
/Roger
From: Michael Kay mike(_at_)saxonica(_dot_)com
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com>
Sent: Saturday, May 9, 2020 9:55 AM
To: xsl-list(_at_)lists(_dot_)mulberrytech(_dot_)com
Subject: [EXT] Re: [xsl] A super-efficient way to compute the sum of A[i] *
B[i]for i=1 to n?
I doubt you'll find much improvement on this.
You could cut out the call on number() and rely on implicit conversion, but I
doubt it makes any difference.
You could factor out the expressions ($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 string-to-double conversion, and
there's no way of avoiding that.
Michael Kay
Saxonica
On 9 May 2020, at 12:59, Costello, Roger L.
mailto:costello(_at_)mitre(_dot_)org
<mailto:xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:
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.
/Roger
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