Yes, if we cache the Delaunay triangulation we could probably do the entire thing in about the time it currently takes to do one time step.

On Thu, May 30, 2019 at 10:50 AM Fernando Paolo notifications@github.com wrote:

@shoyer https://github.com/shoyer and @crusaderky https://github.com/crusaderky That's right, that is how I was actually dealing with this problem prior trying xarray ... by flattening the grid coordinates and performing either gridding (with scipy's griddata) or interpolation (with scipy's map_coordinate) ... instead of performing proper regridding (from cube to cube without having to flatten anything).

As a rule of thumb, any fancy algorithm should first exist for numpy-only data and then potentially it can be wrapped by the xarray library.

This is important information.

For the record, here is so far what I found to be best performant:

import xarray as xr from scipy.interpolate import griddata

Here x/y are dummy 1D coords that wont be used.

da1 = xr.DataArray(cube1, [('t', t_cube1) , ('y', range(cube1.shape[1])), ('x', range(cube1.shape[2]))])

Regrid t_cube1 onto t_cube2 first since time will always map 1 to 1 between cubes.

This operation is very fast.

print('regridding in time ...') cube1 = da1.interp(t=t_cube2).values

Regrid each 2D field (X_cube1/Y_cube1 onto X_cube2/Y_cube2) one at a time

print('regridding in space ...') cube3 = np.full_like(cube2, np.nan) for k in range(t_cube1.shape[0]): print('regridding:', k) cube3[:,:,k] = griddata((X_cube1.ravel(), Y_cube1.ravel()), cube1[k,:,:].ravel(), (X_cube2, Y_cube2), fill_value=np.nan, method='linear')

Performance is not that bad... for ~150 time steps and ~1500 nodes in x and y it takes less than 10-15 min.

I think this can be sped up by computing and saving the interpolation weights between grids in the first iteration and cache them (I think xESMF does this).

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