Added scipy RBF interpolation

[alexknyshu]

The rbf() function is now called from standard scipy library.

[alexknyshu]

Old rbf() function is completely removed.

[alexknyshu]

Import Rbf() function from scipy as rbf().

[alexknyshu]

New rbf() function requires another input format. The input list is transposed and then unpacked with *.

Old: [ [ x1, x2, ... , xd, val ] , [ ... ] , ... ]
New: [ x1, ... ] , [ x2, ... ] , ... , [ xd, ... ] , [ val, ... ]

Original function included cubic basis function, so now it's called manually in arguments.

[paulknysh]

Thanks! I'm actually getting different answers with new and old RBF. Any idea why?

points = np.array([[0., 0., 1.], [0., 1., 1.], [1., 0., 1.], [1., 1., 1.], [0.5, 0.5, 0.]])

test_point = [0.3, 0.3]

old_fit = rbf_old(points)
print(old_fit(test_point))

new_fit = rbf(*np.transpose(points), function='cubic')
print(new_fit(*test_point))

[alexknyshu]

It looks like you have an additional polynomial b^t x + a in the interpolation function

https://arxiv.org/pdf/1605.00998.pdf (equation 4)

while scipy uses only the first "sum-lambda-phi" term by default:

https://github.com/scipy/scipy/blob/adc4f4f7bab120ccfab9383aba272954a0a12fb0/scipy/interpolate/rbf.py#L290

self._function(r) here is your basis function phi(r) and self.nodes are weights lambda, so they simply calculate their dot product and that is it. As far as I see there are no options to extend your interpolation function with any kind of polynomial in scipy.

I'm pretty sure it's true, cause I coded my own rbf without extra terms and get exactly the same results as I get in the standard library.