Extract all dangles of a service area in QGIS

Question

I use the Service area (from layer) tool in QGIS (3.34), which allows to output the boundary nodes. I was expecting that this tool only returns the nodes at all dangles of the calculated service area lines, but I get the dangles plus all other vertices of the service area (as one multipoint geometry). This seems like a bug to me—or I have the wrong understanding of "boundary nodes".

What is the most performant way to only get the service area dangles as nodes?

What I have tried so far

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Tara's solution here (version 1)

Result with visible errors:

Based on this model

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BERA's solution here (version 2)

Same result with errors.

Based on this model

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Babel's solution here (version 3)

All service area vertices remain. Obviously I am still missing something.

Based on this model

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You can find the complete test data here

Answer 1

Try this:

1. Explode lines on Service area layer, to create a line for each line segment.
2. Create a spatial index on the exploded lines.
3. Extract the vertices from the exploded lines.
4. Buffer the vertices with a small distance, I use 5 m.
5. Sum line lengths with the buffer output as Polygons and the indexed layer as Lines layer. Output will be a duplicate of the Buffer layer but with a Count attribute with the number of exploded lines intersecting each buffer.
6. Extract polygons by expression "COUNT"=1, to extract the end points which will only have one line to it.
7. Extract the vertices intersecting the extracted polygons:

My line network have some duplicate lines so all end points arent found. Maybe your line layer is better..:

Answer 2

Due to the problem that service areas (almost) always contain duplicate line segments, it had to be solved like this as an extended version of BERA's answer:

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1. "Explode lines" of service area
2. "Extract vertices" of the exploded lines
3. Create a minimal "Buffer" for the extracted vertices
4. "Sum line lengths" of minimal "Buffer" and result from "Explode lines"
5. "Join attributes by location" (intersect) of "Extract vertices" and "Sum line lengths" results
6. "Extract by expression" (count=1) from result of "Join attributes by location"
7. "Extract by expression" (count=2) from result of "Join attributes by location"

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8. "Extract specific vertices" from original network layer (0,-1 for start point end end point of each line segment)
9. Add a minimal "Buffer" to the extracted network vertices
10. Perform an "Intersection" of the "Extract by expression" count=2 result from above

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11. "Merge vector layers" of the intersection result and the result of "Extract by expression" count=1 result from above.

This is due to the fact that because of overlapping line segments in the calculated service area, all vertices that are corresponding end points and end points in the original network have a count of 2, whereas all service area vertices that have no counterparts in the original network because they area located between its vertices have a count of 1. Thus, the vertices with a count of 2 only have to be kept if they correspond to a start point or an end point of the original network. See the following screenshot (the green count=2 vertices have to be removed).

I can hardly imagine that this is the most performant way to extract all dangles of a service area in QGIS, but so far (and based on the suggestions of others) this is the only working solution that I have found (at least for my specific use case).

Please note: Have a look at

• Remove partly overlapping geometries from service area in QGIS and
• Calculate network length covered by service area in QGIS

as I reason why I'm not satisfied with my own answer and thus can't accept it at the moment.

Answer 3

Here's my final solution:

It is based on:

• BERA's answer and my answer to "Remove partly overlapping geometries from service area in QGIS" (with "Snap geometries to layer" as crucial part of the fix)
• This answer to "Finding number of lines touching intersection point (in polyline layer)"

It seems to me that this is a "cleaner" solution than this one.