David Carlisle wrote:
I expected that the intersection of a and b will produce a count of 2,
ah. In that case you asked the wrong question.
Your node sets are disjoint, no node has both @name='Asm' and @name='Cmp'
so the intersection of the sets $a and $b is empty.
You don't want the intesection of the sets, you want all elements of $a
that have a string value equal to some member of $b, that's easier than
the question you asked,
$a[.=$b]
Err... I think that you mean:
$a[(_at_)obid = $b/@obid]
Right? The standard for "equality" is that the "obid" attributes have the
same value. Your expression will result in "$a" all over again, since all
the elements in "$a" and "$b" have empty string values.
except that if you have the possibility of duplicate obid values you may
need to remove duplicates as well.
Note that
$a[ count( . | $b ) = count( $b ) ]
is the same as
$b[ count( . | $a ) = count( $a ) ]
it really is the nodes in both $a and $b,
Well, except possibly for order, as I noted earlier.
but
$a[.=$b]
$a[(_at_)obid = $b/@obid]
is not the same
as
$b[.=$a]
$b[(_at_)obid = $a/@obid]
one is the set of nodes in $a the other is the set in $b.
Right, in the first expression all the "Class" elements will have a "name"
attribute with the value "Asm". In the second expression all the "Class"
elements will have a "name" attribute with the value "Cmp".
-- Roger Glover
glover_roger(_at_)yahoo(_dot_)com
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