On Tue, Jul 31, 2012 at 11:16 AM, Daniel Wheeler <[email protected]>wrote:
> On Mon, Jul 30, 2012 at 10:58 PM, Daniel Farrell <[email protected]>wrote: > >> >> > Just use what you have written. Should work as is. Actually, it should be > > numpy.exp( -scipy.quad( lambda x: phi([[x]]), 0, x)[0] ) > > if phi is a cell variable. The __call__ method of cell variable is used to > pass in an array of positions of shape (dim, numberOfPositions) and then > the returned array with shape (numberOfPositions,) is the interpolated > value of the cell variable at the given positions. > This might be crazy slow if this is done point by point. Might be better to create the interpolated phi array beforehand in one shot and then do the integral. -- Daniel Wheeler
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