I will try this as well and report back with a timing... On Mon, Jan 24, 2011 at 3:56 PM, Vincent Schut <sc...@sarvision.nl> wrote: > On 01/24/2011 02:53 PM, John wrote: >> Hello, >> >> I'm trying to cycle over some vectors (lat,lon,emissions) of >> irregularly spaced lat/lon spots, and values. I need to sum the values >> each contributing to grid on a regular lat lon grid. >> >> This is what I have presently, but it is too slow. Is there a more >> efficient way to do this? I would prefer not to create an external >> module (f2py, cython) unless there is really no way to make this more >> efficient... it's the looping through the grid I guess that takes so >> long. > > Use np.histogram2d with weights=emissions, and lat and lon as your x and > y to histogram. Choose the bins to match your grid, and it will > effectively sum the emission values for each grid cell. > > Vincent. > >> >> Thanks, >> john >> >> >> >> def grid_emissions(lon,lat,emissions,grid.dx, grid.dy, >> grid.outlat0, grid.outlon0, grid.nxmax, grid.nymax): >> """ sample the emissions into a grid to fold into model output >> """ >> >> dx = grid.dxout >> dy = grid.dyout >> >> # Generate a regular grid to fill with the sum of emissions >> xi = np.linspace(grid.outlon0, >> grid.outlon0+(grid.nxmax*grid.d), grid.nxmax) >> yi = np.linspace(grid.outlat0, >> grid.outlat0+(grid.nymax*grid.dy), grid.nymax) >> >> X, Y = np.meshgrid(yi, xi) >> Z = np.zeros(X.shape) >> >> for i,x in enumerate(xi): >> for j,y in enumerate(yi): >> Z[i,j] = np.sum( emissions[\ >> np.where(((lat>y-dy)& (lat<y+dy))& >> ((lon>x-dx)& (lon<x+dx)))[0]]) >> >> return Z > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >
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