New answers tagged spatial-statistics
A quick google search yielded this result which states in the paragraph before "Continents and Oceans" that 'this land hemisphere is centered on Western Europe.' You could validate this and get the precise location with functions in PostGIS by finding the centroid of a polygon (landmass). Then you can do some math to find the cut line based on the centroid.
One option would be to look at a multivariate regression in R. Let us assume that you have create a data.frame with your density, lat & lon. Then you can do something along these lines (rough outline taken from Quick-R): # Multiple Linear Regression fit <- lm(density ~ lat + lon, data=DensityDataFrame) summary(fit) # show results # Other useful ...
Just realised that it's simply the same process as in older mapinfo versions ... so Sum(ObjectLen(obj,"m"))"Length_m" Thanks to http://web.pb.com/mapinfopro-oct-2013/Query-Ninja-Calculating-Line-Objects-Length/?webSyncID=46d43e17-2562-4890-a3d6-aae32714dc02
Yes, you can. You need to use the Geostatical Analyst's "Areal Interpolation" method in ArcGIS. This will give you a statistically sound "smoothed" / interpolated surface for your data.
It is a pure Python problem With shapely, your 3D point is represented by pt3D = Point(11.52951677300007,0.7729100360000416, -50000) list(pt3D.coords) [(11.52951677300007, 0.7729100360000416, -50000.0)] Using slicing for example (there are others solutions) pt2D = Point(list(pt3D.coords)[:2]) print pt2D POINT (11.52951677300007 0.7729100360000416) ...
Remove overlaps and zonal statistics will perform as expected. Extract from tool help: If the zone feature input has overlapping polygons, the zonal analysis will not be performed for each individual polygon. Since the feature input is converted to a raster, each location can only have one value. An alternative method is to process the zonal ...
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