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pythonscipy
Avalokitesvara
Avalokitesvara

Reputation: 97

Evaluating derivative of a spline using splev in Scipy

I have created a bspline using splprep as below from a set of points:

tck,uout = splprep([x,y],s=0.,k=2,per=False)

Now, I am trying to evaluate the derivative of a spline using:

dx,dy = splev(uout,tck,der=1)

I find that splev returns two lists for the derivative.

Given that the Spline is parametrized (say in u), does it return dx/du and dy/du ?

If not how to evaluate the derivative (dy/dx) properly ?

Upvotes: 5

Views: 3750

Answers (1)

James K
James K

Reputation: 3752

Yes, if der = 1 the the lists are the values of dx/du and dy/du at each point. The gradient is then dy/dx = dy/du / dx/du.

I'm slightly concerned about the splprep call: s is optional, but if defined it should have a value of about the same as the number of points (larger means smoother). per is an integer value, not a boolean. And cubic splines are better behaved than quadratic. http://docs.scipy.org/doc/scipy/reference/generated/scipy.interpolate.splprep.html

Upvotes: 4

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