## Re: [xsl] Increasing sequence ?

 Subject: Re: [xsl] Increasing sequence ? From: "Dimitre Novatchev dnovatchev@xxxxxxxxx" Date: Wed, 25 Mar 2015 17:36:24 -0000
```> every \$v in 1 to (count(\$sequence)-1) satisfies (\$sequence[\$v] lt \$sequence[\$v+1])

This may be inefficient (depending on the XPath engine being used).

I think the following has better chances of being more efficient --
write an xsl:function, name it, say, "increasing". The body of this
function can be just:

not(\$seq[2])  or \$seq[1] lt \$seq[2]  and  increasing(subsequence(\$seq, 2))

Cheers,
Dimitre

On Wed, Mar 25, 2015 at 10:20 AM, Leo Studer leo.studer@xxxxxxxxxxx
<xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx> wrote:
> A similar problem as before, is the integer sequence increasing?
>
> this is my solution:
>
> every \$v in 1 to (count(\$sequence)-1) satisfies (\$sequence[\$v] lt \$sequence[\$v+1])
>
> Do you have a better one?
>
> Cheers
> Leo
>

--
Cheers,
Dimitre Novatchev
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