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4

For general centroid creation, QGIS QgsGeometry functions wrap respective calls to the underlying GEOS library: QgsGeometry * QgsGeometry::centroid() (definition) calls GEOSGetCentroid (definition) QgsGeometry * QgsGeometry::pointOnSurface() (definition) calls GEOSPointOnSurface (definition) The pointOnSurface() function is new for QGIS 2.4 and allows a ...


4

To do this I would use the Intersect (Analysis) tool. The ArcPy code will be: import arcpy homesFC = <your path to input Homes> districtsFC = <your path to input Districts> homesWithDistrictFC = <your path to output> arcpy.Intersect_analysis ([homesFC, districtsFC], homesWithDistrictFC)


3

For the Points that fall within the 65 meter radius, can't you just subtract 5 meters from A's Y coordinate and use A's X coordinate to find a Point at the 5 meter position? You can use the DA cursor to update these Point's Geometry to the 5m position.


3

If you have access to the 3D Analyst extension, this can be accomplished using the TIN Node tool found here: 3D Analyst Tools - Conversion - From TIN - TIN Node. This will create 3D points (i.e Z-enabled) at each triangle node that forms the TIN. Alternatively you could convert your TIN to a raster, and then use the Raster to Multipoint tool which will ...


2

You can also visualize this by clicking your line symbology in the Table of Contents. Under the Esri line styles, select the symbology called "Arrow at End" and this will show you the directionality of the line.


1

You could use the Distance to nearest hub algorithm from the Processing plugin which you could add your points and your lines layer instead of going through a query. From a couple of example layers that I have, you can get something like this: Then when you open up the Attributes Table for the output layer, you will be given the HubName and HubDist in ...


1

If your goal is to SEE where the line ends, you can also add an arrow symbol at the end of your line. (layer properties > symbology >symbol properties > cartographic line symbol > line properties). Otherwise, for a non programmatic solution, you can compute the X and Y coordinates of the end point based on "calculate geometry"(right click on field), then ...


1

This looks familiar! The issue you're having is that the result of cent is a multipoint geometry instead of just a point geometry- even though the result of the intersect IS a single point. I'm not sure if a multipoint is always returned or what. You might try something like this: T2B = arcpy.Array([arcpy.Point(1,2),arcpy.Point(-1, -2)]) L2R = ...


1

How about this for an idea. If B is within the specified distance of A then create a polyline between A and B. You can then create a point any distance along that polyline. Polyline object has a method positionalongline.



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