[Numpy-discussion] need a better way to fill a grid

David Huard david.huard at gmail.com
Mon Jan 24 09:50:35 EST 2011


Hi John,

Since you have a regular grid, you should be able to find the x and y
indices without np.where, ie something like

I = (lon-grid.outlon0 / grid.dx).astype(int)
J = (lat-grid.outlat0 / grid.dy).astype(int)

for i, j, e in zip(I, J, emissions):
    Z[i,j] += e


David

On Mon, Jan 24, 2011 at 8:53 AM, John <washakie at gmail.com> 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.
>
> 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
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