spatial: How to get indexed nodes from LayerNodeIndex?

[Axel Morgner]

Hi,

I'm struggling with getting the indexed node from a LayerNodeIndex.

Let's say adding a node (id 1) to the layer index:

ￜￜￜ layerIndex.add(node, "", "");

In LayerNodeIndex, the id is saved as id property on the SpatialDatabaseRecord object.

ￜￜￜ layer.add(
ￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜ decodeGeometry, new String[] { "id" },
ￜￜￜￜￜￜￜￜￜￜￜￜￜￜￜ new Object[] { geometry.getId() } );

Internally, (in EditableLayerImpl), an additional geomNode (let's say id 2) is created:

ￜￜￜ private Node addGeomNode(Geometry geom, String[] fieldsName, Object[] fields) {
ￜￜￜ ￜￜￜ Node geomNode = getDatabase().createNode();
ￜￜￜ [...]

When querying the index, the iterator returns node 2, not node 1.

SpatialRecordHits:

ￜￜￜ protected Node fetchNextOrNull()
ￜￜￜ {
ￜￜￜￜￜￜￜ int i = index++;
ￜￜￜￜￜￜￜ return i < size() ? hits.get( i ).getGeomNode() : null;
ￜￜￜ }

I would expect to get the original node (id 1 here) from the index. Am I missing something?

Greetings
Axel

[Peter Neubauer]

Well,
Thing is that there might not be a one to one mapping between geometry and your domain node. It might just be the entry point to a whole polygon or a traversal. That is why you get back the geometry node, and the decoder can then resolve it to original underlying domain subgraphs (in the easiest case only one node)

Makes sense?

On May 3, 2012 6:16 PM, "Axel Morgner" <ax...@morgner.de> wrote:

Hi,

I'm struggling with getting the indexed node from a LayerNodeIndex.

Let's say adding a node (id 1) to the layer index:

    layerIndex.add(node, "", "");

In LayerNodeIndex, the id is saved as id property on the SpatialDatabaseRecord object.

    layer.add(

                decodeGeometry, new String[] { "id" },

                new Object[] { geometry.getId() } );

Internally, (in EditableLayerImpl), an additional geomNode (let's say id 2) is created:

    private Node addGeomNode(Geometry geom, String[] fieldsName, Object[] fields) {

        Node geomNode = getDatabase().createNode();
    [...]

When querying the index, the iterator returns node 2, not node 1.

SpatialRecordHits:

    protected Node fetchNextOrNull()
    {
        int i = index++;

        return i < size() ? hits.get( i ).getGeomNode() : null;
    }

[Axel Morgner]

Thanks Peter for the quick reply!

Ok, with a decoder (f.e. the SimplePointEncoder#decodeGeometry), I get the geometry, but still no reference to my domain node which I added to the index.

Solved it for me by using graphDb.getNodeById((Long) geomNode.getProperty("id")).

[Peter Neubauer]

That is true. Do you think that would be better? Craig, WDYT of
changing this to be a relationship?

Cheers,

/peter neubauer

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L:   http://www.linkedin.com/in/neubauer
T:   @peterneubauer

If you can write, you can code - @coderdojomalmo
If you can sketch, you can use a graph database - @neo4j

[Craig Taverner]

This differs somewhat from my memory of how things work. In the SimplePointLayer the node added to the index is in fact the geometry node. Are you using some other GeometryEncoder that expects the node to be something else?

On the other hand, reading the code above I see you are actually not working with the original index, but the LayerNodeIndex which Peter added as a wrapper. In that case I think only Peter knows the reason for not storing and retrieving the node originally passed. I know very little about the indexProvider wrappings.

My own vote is to keep to the original model conceptually, which is that whatever you add to the index is whatever you get back, and in principle this should be the node that represents the geometry. How that geometry is actually stored is up to the geometry encoder, and can be separate from the node passed, but is usually either the node passed (as in SimplePointGeometryEncoder and DefaultGeometryEncoder) or closely linked by relationships to it (as in the OSMGeometryEncoder).

Peter, do you remember why you used a different node in the IndexProvider?

[Peter Neubauer]

No, I don't. Thisight actually be a bug. Axel, if you change it and send a PR ad I am not on my machine, I'd that OK?

[Axel Morgner]

Ok! If I find a little time tomorrow, I will have a look on it and make a suggestion and PR.

[Peter Neubauer]

You rock.

[Axel Morgner]

Unfortunately, I did not succeed in making changes that won't break the tests. We'll see what insights tomorrow will bring perhaps ..

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