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Apr 28, 2020 at 16:05 comment added Abdirizak Ok, I get the point. I should be able to handle the large rasters in some way. Thanks
Apr 28, 2020 at 10:34 comment added Spacedman You can estimate the spatial range from the variogram, and you can produce that from subsets of the locations if you have too many grid points to do the whole thing. I used autokrige in my answer because its an all-in-one solution and scaling to large grids wasn't mentioned as an issue.
Apr 28, 2020 at 10:32 comment added Spacedman I mean the spatial range of the correlation. If you have a 1000x1000 1meter grid and your spatial correlation range is about 100m then the amount of information got from two adjacent grid cells is going to be not much different from one cell. Hence if computing with the 1000000 cells is a problem you can sample a subset of locations and not do much worse with your predictions.
Apr 28, 2020 at 9:51 comment added Abdirizak Ok. But what do you mean by the correlation being larger than the raster cell size? Correlation ranges from -1 to +1, whereas the cell size can be anything in meters. I don't think the two are readily comparable. Or is there some formula that sets them into relation?
Apr 25, 2020 at 12:49 comment added Spacedman If you have a large raster and your spatial correlation is a lot bigger than the raster cell size then you can use a sample of the known points and you shouldn't get a much worse answer. Random thinning or maybe keeping more points near your unknown points...Hmmm
Apr 25, 2020 at 12:14 vote accept Abdirizak
Apr 25, 2020 at 12:13 comment added Abdirizak Thanks a lot for your help! I made one small adjustment: after xyV = as.data.frame(r,xy=TRUE), I added: colnames(xyV) <- c("x", "y", "layer"). With this, your code runs for me. It seems to be quite slow on large rasters though, my testing suggests because of the automap::autoKrige function (i.e. the Kriging).
Apr 24, 2020 at 16:05 history answered Spacedman CC BY-SA 4.0