Geographic Information Systems Asked by Pablo Herreros Cantis on April 16, 2021
My ultimate goal is to convert landcover raster (.tif) objects to an sf object representing the raster’s grid and the original values of each cell within each geometry. I have been able to do this for smaller rasters doing the following:
library(sf)
library(stars)
# import raster using stars
landcover_stars <- read_stars(my_raster.tif)
# convert to sf object using st_as_sf
landcover_grid_sf <- st_as_sf(landcover_stars)
In larger rasters (e.g. my largest raster is currently 11482×12607 cells), however, the read_stars() function imports the raster as a "stars proxy", which is a step taken to handle large raster datasets by the package. While stars proxy objects are not accepted by the st_as_sf
function, it is possible to set "proxy = FALSE" in the function. If I do this in my largest dataset, however, running st_as_sf(landcover_stars)
with the resulting object will crash my laptop {16 GB RAM, i7 2.70GHz processor}.
Is there a way I can proceed to ease the load on my machine when converting very large star objects to sf?
In addition – could it be that it is actually the newly generated sf object what is depleting my machine?
Here is a dummy raster in case youd like to test it, with integer values randomly generated ranging from 1 to 10:
raster(nrows=12000, ncols=12000, xmn=0, xmx=10, vals = floor(runif(12000*12000, min=0, max=11)))
The stars
package (latest version) has it's own version of st_as_sf
with two parameters that might help here: merge = TRUE
and as_points = FALSE
. So you might try:
stars::st_as_sf(landcover_stars, merge = TRUE, as_points = FALSE)
BTW, your example with random distribution of values will not be so helpful since I expect that there are very few clusters of equal values, as you might expect in a real landcover raster. There will be almost 144 million separate tiny polygons.
Answered by Micha on April 16, 2021
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