On Fri, Sep 20, 2013 at 5:38 PM, Markus Metz <markus.metz.gisw...@gmail.com> wrote: > Glynn Clements wrote: >> >> Luca Delucchi wrote: >> >>> maybe v.median [0] could help? >> >> Not for large datasets. First, it requires that the data will fit into >> RAM. Second, numpy.median() sorts the entire array and takes the >> middle value, which is somewhere between O(n.log(n)) for the typical >> case and O(n^2) for the worst case (numpy.median uses the default >> sorting algorithm, which is quicksort). >> >> See r.quantile for a more efficient approach for large datasets. The >> first pass computes a histogram which allows upper and lower bounds to >> be determined for each quantile. The second pass extracts values which >> lie within those bounds and sorts them. > > The approach of v.median and r.quantile regards each dimension > separately which is IMHO not correct in a spatial context.
This > The median > of a point cloud would be a multivariate median for 2 or 3 dimensions. > You would need to sort the points first by the first dimension, then > by the second dimension etc, then pick the coordinates of the mid > point. is wrong, please ignore. This > Alternatively you would need to find the point with the > smallest distance to all other points, which is nightmare to calculate > ( (n - 1) x (n - 1) distance calculations). is correct, but instead of finding the existing point with the smallest distance, that point can be approximated with less effort. Markus M _______________________________________________ grass-dev mailing list grass-dev@lists.osgeo.org http://lists.osgeo.org/mailman/listinfo/grass-dev