The best I could come up with was
tail(fold-left($x, 0, function($a, $b) {$a, $a[last()] + $b}))
Your solution could be rewritten
fold-left($x, (), function($a, $b) {$a, ($a[last()], 0)[1] + $b})
Another option is
fold-left(tail($x), $x[1], function($a, $b) {$a, $a[last()] + $b})
xsl:iterate is more intuitive though of course more verbose:
<xsl:iterate select="1 to 4">
<xsl:param name="total" select="0"/>
<xsl:variable name="new-total" select="$total + ."/>
<xsl:sequence select="$new-total"/>
<xsl:next-iteration>
<xsl:with-param name="total" select="$new-total"/>
</xsl:next-iteration>
</xsl:iterate>
Michael Kay
Saxonica
On 7 Mar 2019, at 13:28, Martin Honnen martin(_dot_)honnen(_at_)gmx(_dot_)de
<xsl-list-service(_at_)lists(_dot_)mulberrytech(_dot_)com> wrote:
Am 07.03.2019 um 12:55 schrieb Michael Kay mike(_at_)saxonica(_dot_)com:
A good simple use case for fold-left() is to accumulate a running total,
i.e. turn (1,2,3,4) into (1,3,6,10).
The example to simply compute the running total (e.g. map (1,2,3,4) to 10) is
in the spec with
fold-left((1 to 4), 0, function($a, $b) { $a + $b})
But to map the whole sequence (1,2,3,4) with fold-left to a new sequence of
(1,3,6,10) I am already struggling to express that in a compact way, is
fold-left(
(1 to 4),
(),
function ($a, $b) { $a, if (empty($a)) then $b else $b + $a[last()] }
)
a good way? Or can the third argument, the function be expressed in a more
compact way?
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