## Python (numpy) to Matlab/Octave conversion problem

Started by 2 years ago●4 replies●latest reply 2 years ago●177 viewsI'm new to python language. I can't figure out what happens here in code:

def evaluate(self, x): p1 = 1 p2 = np.zeros((1, self.r), dtype=self.dtype) p2 += self.coeff[0, np.newaxis, :] for k in range(1, self.n): v = x - self.xcoord[k-1] p1 = v*p1 p2 += p1[:, np.newaxis] * self.coeff[k] return p2 where, n = array size of xcoord r = self.ycoord.shape coeff = np.zeros((self.n+1, self.r), dtype=self.dtype)

How these

self.coeff[0, np.newaxis, :]

and

p1[:, np.newaxis]

can be converted to Matlab/Octave language ?

p1 is defined as variable but, then later it is used as an array type variable. coeff is an array type variable ... but, what is done in command

self.coeff[0, np.newaxis, :]

?

In NumPy,

np.newaxis

is used to add 'dimension' to the vector or matrix.

The * is used for one-to-one multiplication of elements (not matrix product). I guess 'np.newaxis' is used to have matching matrix dimensions

Yes, it looks like np.newaxis does not (in these cases) change the dimensions so, the function can be converted to Matlab/Octave language as like:

% evaluation

p1 = 1;

p2 = zeros(1, 1);

p2 += coeff(1);

for k = 1:n

v = x - xcoord(k);

p1 = v*p1;

p2 += p1 * coeff(k+1);

endfor

Hi,

`p2 += p1[:, np.newaxis] * self.coeff[k]`

means

`p2 = p2 + p1[:, np.newaxis] * self.coeff[k]`

I'm C++ developer mainly so that is std syntax ... I'll edit the question a bit.