If you have the latest version of geopandas (0.3.0 as of writing), and the if df is a GeoDataFrame, you can use the x and y attributes on the geometry column: df[‘lon’] = df.point_object.x df[‘lat’] = df.point_object.y In general, if you have a column of shapely objects, you can also use apply to do what … Read more
While the answer of eguaio does the job, there is a more natural way to get the closest point using shapely.ops.nearest_points function: from shapely.geometry import Point, Polygon from shapely.ops import nearest_points poly = Polygon([(0, 0), (2, 8), (14, 10), (6, 1)]) point = Point(12, 4) # The points are returned in the same order as … Read more
If you specifically want to construct your Polygon from the shapely geometry Points, then call their x, y properties in a list comprehension. In other words: from shapely import geometry poly = geometry.Polygon([[p.x, p.y] for p in pointList]) print(poly.wkt) # prints: ‘POLYGON ((0 0, 1 0, 1 1, 0 1, 0 0))’ Note that shapely … Read more
The GIS term you are describing is linear referencing, and Shapely has these methods. # Length along line that is closest to the point print(line.project(p)) # Now combine with interpolated point on line p2 = line.interpolate(line.project(p)) print(p2) # POINT (5 7) An alternative method is to use nearest_points: from shapely.ops import nearest_points p2 = nearest_points(line, … Read more
Shapely can directly parse this: import shapely.wkt P = shapely.wkt.loads(‘POLYGON ((51.0 3.0, 51.3 3.61, 51.3 3.0, 51.0 3.0))’) print(P)
This type of intersection is easily done by the “min of the maxes” and “max of the mins” idea. To write it out one needs a specific notion for the rectangle, and, just to make things clear I’ll use a namedtuple: from collections import namedtuple Rectangle = namedtuple(‘Rectangle’, ‘xmin ymin xmax ymax’) ra = Rectangle(3., … Read more
Consider using Rtree to help identify which grid cells that a polygon may intersect. This way, you can remove the for loop used with the array of lat/lons, which is probably the slow part. Structure your code something like this: from shapely.ops import cascaded_union from rtree import index idx = index.Index() # Populate R-tree index … Read more
I found a solution that works for the specific case given: >>> pp2 = pp.buffer(0) >>> pp2.is_valid True >>> pp2.exterior.coords[:] [(0.0, 0.0), (0.0, 3.0), (3.0, 3.0), (3.0, 0.0), (2.0, 0.0), (0.0, 0.0)] >>> pp2.interiors[0].coords[:] [(2.0, 1.0), (2.0, 2.0), (1.0, 2.0), (1.0, 1.0), (2.0, 1.0)]
Here is a possible solution to my problem. Geographical coordinates must be stored properly. Example np.array([[Lon_A, Lat_A], [Lon_B, Lat_B], [Lon_C, Lat_C]]) Create the polygon Create the point to be tested Use polygon.contains(point) to test if point is inside (True) or outside (False) the polygon. Here is the missing part of the code: from shapely.geometry import … Read more
Note: cascaded_union mentioned in the answer below is superceded by unary_union if GEOS 3.2+ is used – this allows unions on different geometry types, not only polygons. To check your version, >>> shapely.geos.geos_version (3, 5, 1) From the question/answer here, it seems this is called a cascaded_union within shapely: from shapely.ops import cascaded_union polygons = … Read more