Hi Thomas,
Could you describe your graph better?

Lvc@


On 8 March 2015 at 16:22, Thomas Müller <[email protected]>
wrote:

> Hello,
>
> for testing i generated a category hierachy ("V_Category") which has 4
> levels and 5 categories per level, starting from one ROOT-category (#17:0).
> The hierachy is build up with "E_CH" category hierachy edges (
> parentCategory.addEdge(EEdges.E_CH.name(), childCategory); ). At all 1364
> categories.
>
> I also generated 500 documents which have assigned randomly 1 to 10 of the
> above categories:
>
>
>        nextInt = RandomUtils.nextInt(9) + 1;
>         for (int j = 0; j < nextInt; j++) {
>             OrientVertex orientVertex = nextRandomCategory().
> getOrientVertex();
>             document.addEdge(EEdges.E_Category.name(), orientVertex);
>         }
>
>
>
> i want to select the number of documents assigned to a category. What i
> tried ist:
>
> select count(in('E_Category')) from (traverse out('E_CH'), in('E_Category'
> ) from #17:0)
>
> What i expect to get is 500 since i have 500 documets. What i get is 1866
> which is 1364(number of categories) +500(number of documents) +2 (?). Even
> if i change the number of categories this formula seems to be constant:
> number of categories+number of documents+2. The good thing: the duplicates
> (one cocument may have several categories) already seem to be removed (no
> distinct necessary).
>
> How do i have to change the querry to get only the number of documents
> which have assigned one or more categories from the given root category and
> all child categories.
>
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