currently the algorithm returns only the vertices , but how to get the edges that it chose ? as there can be multiple edges between the vertices
I'll have to modify current function. Search source code 'OSQLFunctionDijkstra.java' for existing function and inside 'getDistance' method it should be easy to print 'Edge' information.
Related
I am stuck at a current project:
I have an input picture showing the ground with some shapes on it. I have to find a specific shape with a given template.
I have to use distance transformation into skeletonization. My question now is: How can I compare two skeletons? As far as I noticed and have been told, the most methods from the Image Processing Toolbox to match templates don't work, since they are not scale-invariant and rotation invariant.
Also some skeletons are really showing the shapes, others are just one or two short lines, with which I couldn't identify the shapes, if I didn't know what they should be.
I've used edge detection, and region growing on the input so there are only interessting shapes left.
On the template I used distance transformation and skeletonization.
Really looking forward to some tips.
Greetings :)
You could look into convolutions?
Basically move your template over your image and see if there is a match, and where.
The max value of your array [x,y] is the location of your object in the image.
Matlab has a built-in 2D convolution function for this
I create a network add nodes and edges. I view it (it creates a dot and pdf file automatically). Later, I want to create a second network with the same nodes but different edges. I want to place the nodes in the same coordinates, so that I can make a comparison of both graphs easily. I tried to get the coordinates of the first graph, and tried to set the coordinates of the nodes) but I couldn't find proper functions to do that. I also checked networkx package. I also tried to get a copy of the first network, and delete the edges with no success. Can someone please show me how to create a second network with the same node coordinates?
This is the simple network creation code
import graphviz as G
network1 = G.Digraph(
graph_attr={...},
node_attr={...},
edge_attr={...} )
network.node("xxx")
network.node("yyy")
network.node("zzz")
network.edge("xxx", "yyy")
network.edge("yyy", "zzz")
network1.view(file_name)
First, calculate the node positions for the first graph using the layout of your choice (say, the spring layout):
node_positions = nx.layout.spring_layout(G1)
Now, you can draw this graph and any other graph with the same nodes in the same positions:
nx.draw(G1, with_labels=True, pos=node_positions)
nx.draw(G2, with_labels=True, pos=node_positions)
Graphviz's layers feature might also be interesting:
https://graphviz.org/faq/#FaqOverlays
Here is a working example of using layers - ignore the last two lines that create a video.
https://forum.graphviz.org/t/stupid-dot-tricks-2-making-a-video/109
And here is some more background:
https://forum.graphviz.org/t/getting-layers-to-work-with-svg/107
I think i have a difficult problem right here..
I want to able to get the surfaces of f.e. the orange object in this three.js example https://threejs.org/examples/?q=stl#webgl_loader_stl
i want to click with the mouse, find the correct surface, which should then be highlighted, so i make sure this was the surface i want.
(i already implemented raycaster successfully, so thats not an issue)
The intersectObject method returns an array of intersections, each of which has face property. The face contains vertex indices.
For STL files containing multiple solids, each solid is assigned to a different group, and the groups are available in the geometry object that is returned from STLLoader. Each group is defined by a range of vertex indices.
So, I think you can correlate the vertex indices returned from the raycaster with the vertex indices in the geometry groups.
There is a script Here that can process the data from OSM to detect intersections in a given bounding box. It works by getting all the ways in a given bounding box and then finding other ways that share common nodes with these roads. Here is the query that does that,
way
["highway"]
["highway"!~"footway|cycleway|path|service|track|proposed"]
(, , , )
->.relevant_ways;
foreach.relevant_ways->.this_way{
node(w.this_way)->.this_ways_nodes;
way(bn.this_ways_nodes)->.linked_ways;
way.linked_ways
["highway"]
["highway"!~"footway|cycleway|path|service|track|proposed"]
->.linked_ways;
(
.linked_ways->.linked_ways;
-
.this_way->.this_way;
)->.linked_ways_only;
node(w.linked_ways_only)->.linked_ways_only_nodes;
node.linked_ways_only_nodes.this_ways_nodes;
out;
}
This query returns all kinds of intersections (4-way intersections, T-junctions, ... ).
Question:
Is there a way to further filter out the intersection to 4-way intersections and T-junctions?
One idea I have is to check if the common node is an endpoint of one the ways, which makes the intersection and if the common node is in the middle of the road it will be a 4-way intersection. But I am not sure how to write a query that does this.
Any help will be appreciated, Thanks.
I would like to find all highway way member nodes in a certain radius. I cannot see how to do this without using intersection, however, that is not in the API. For example I have this:
[out:json];
way(around:25, 50.61193,-4.68711)["highway"];>->.a;
(node(around:25, 50.61193,-4.68711) - .a);
out;
Result set .a contains the nodes I want but also nodes outside the radius - potentially a large number if the ways are long. I can find all the nodes inside the radius I don't need, as returned by the complete query above. Now I can always perform a second around query and do the intersection of the two result sets outside of Overpass. Or I can do another difference:
[out:json];
way(around:25, 50.61193,-4.68711)["highway"];>->.a;
(node(around:25, 50.61193,-4.68711) - .a)->.b;
(node(around:25, 50.61193,-4.68711) - .b);
out;
This gives the result I want but can it be simplified? I'm certain I'm missing something here.
Indeed, your query can be simplified to an extent that we don't need any difference operator at all. I would recommend the following approach:
We first query for all nodes around a certain lat/lon position and a given radius.
Based on this set of nodes we determine all ways, which contain some of the previously found nodes (-> Hint: that's why we don't need any kind of intersection or difference!).
Using our set of highway ways we now look again for all nodes of those ways within a certain radius of our lat/lon position.
In Overpass QL this reads like:
[out:json];
node(around:25, 50.61193,-4.68711);
way(bn)[highway];
node(w)(around:25, 50.61193,-4.68711);
out;
Try it on Overpass Turbo