[xsl] exercise in complex grouping

Subject: [xsl] exercise in complex grouping
From: "Syd Bauman s.bauman@xxxxxxxxxxxxxxxx" <xsl-list-service@xxxxxxxxxxxxxxxxxxxxxx>
Date: Tue, 12 May 2020 09:33:42 -0000
I have a moderately sizable TEI file (~31,000 text nodes with ~100,400
"words" or ~688,000 characters; ~20,000 elements, ~15,000 attributes).
Somewhere in all that mess there are a few pairs of elements for which
I need some special processing.

Say each pair is an <A> and a <B>. I can find each <B> by XPath quite
trivially. In addition, for every pair, <B> has a @target that points
to the corresponding <A> via a bare name identifier URL. Furthermore,
every <B> in the document is part of such a pair. (Which is why it is
so trivial to find them via XPath. The same can not be said for <A>:
there are *lots* of <A> elements that are not part of an <A>-<B>
pair; but none, of course, that bear that particular @xml:id, so they
can be found by XPath. It's just easy, not trivial. :-)

In general, there can be other nodes between <A> and <B>, and there
will be cases in which <B> precedes rather than follows the <A> it
points to. E.g.,

   blah blah blah
   <d><e>blah</e> blah
   <B target="#A1">blort</B>
   <f>monkey</f> shines
   <A xml:id="A1">snort</A>

I want to be able to handle these cases, too.

For the foreseeable future, there will never be another <B> in between
a <B> and the <A> it points to, and each <B> will be a child of the
same element as the <A> it points to. (I.e., no overlap problems.) But
as soon as I say these complications will never happen, the very next
day the editors will gleeful send e-mail saying they have found such a
case. But for now, if needed, I'm willing to write code that presumes
it won't happen.

What I want for output is to be able to wrap the <B> with the <A> it
points to, *and everything in between* in a <C>.

   blah blah blah
   <d><e>blah</e> blah
   <C xml:id="A1Container">
     <B target="#A1">blort</B>
     <f>monkey</f> shines
     <A xml:id="A1">snort</A>
I am 90% confident I can write some messy XSLT 1.0 Muenchian grouping
code that does this. (Although I suspect it would take two passes,
one for <A> precedes <B>, another for <B> precedes <A>; but I don't
care about two passes at all, and would not even care if it took N
passes.[1]) But I am equally confident there is a much better
<xsl:for-each-group> method that, at the moment, I simply can't wrap
my head around.

Thanks for any thoughts, pointers, code, or advice.

[1] Where N is proportional to the number <A>-<B> pairs.

 Syd Bauman, NRP  (he/him/his)
 Senior XML Programmer/Analyst
 Northeastern University Women Writers Project
 s.bauman@xxxxxxxxxxxxxxxx or

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