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we want to use matlab's msvcr. in shapely, replacing CDLL(find_library('c')) with CDLL('C:\\Program Files\\MATLAB\\R2015a\\bin\\win64\\msvcr100.dll') makes it all work. this answer seems to be a method that would determine that path automatically, appropriate for wherever you are invoking shapely from.


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I assume that the polyline is unbounded; that is, either the polyline comprises a single line or the first and last pieces are rays. Otherwise, the problem is not well defined. You can use the CGAL 2D-Arrangement data structure to solve your problem easily: Construct a 2D-Arrangement induced by the polyline and issue two point locations queries for A and B, ...


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Please check isLeft() function in this link.


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My idea is to make line from A and B points (AB). And Find the crossing points AB and L. If 0 than A and B on the same side, if 1 they aren't. If you have more than one it means A and B on the same side just like in the first case. (It may can happen sometimes AB above Lp1 or below Lpn- so it would be better to extend L endpoints with a distance n) Another ...


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My idea may not be most efficient but anyway, I believe you can at least get a correct answer by creating offset lines to the left and to the right from linestring L and computing shortest distances from A and B to the offset lines. If both A and B have shorter distance to the same offset line it means that they are on the same side of L. That's not true if ...



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