python,numpy,scipy,interpolation,cubic-spline

This is not possible with interp1d. You can get the spline coefficients from splrep though.

python,scipy,interpolation,spline,cubic-spline

Short answer: import numpy as np from scipy import interpolate def f(x): x_points = [ 0, 1, 2, 3, 4, 5] y_points = [12,14,22,39,58,77] tck = interpolate.splrep(x_points, y_points) return interpolate.splev(x, tck) print f(1.25) Long answer: scipy separates the steps involved in spline interpolation into two operations, most likely for computational...

python,math,numpy,spline,cubic-spline

I just found something really interesting with the answer that I need with a bézier in this link. Then I used the code to try on my own. It's working fine apparently. This is my implementation: #! /usr/bin/python # -*- coding: utf-8 -*- import numpy as np import matplotlib.pyplot as...

javascript,algorithm,curve,bezier,cubic-spline

It's Catmull-Rom fitting: the code tries to find an appropriate tangent through point X, based on the location of points X-1 and X+1, such that the tangent is parallel to the line (X-1)--(X+1), and then fiddles with the control points that yields, to make sure the "incoming" and "outgoing" tangents...

matlab,spline,piecewise,cubic-spline

This is probably the easiest way to get pp1 + pp2 Adding to the code in the question: pp12 = @(x) ppval(pp1,x)+ppval(pp2,x); breaks = unique([pp1.breaks,pp2.breaks]); pp3 = spline(breaks,pp12(breaks)); plot(tnew,ppval(pp3,tnew),'k:'); Gives the dotted black line: with pp3 being in piecewise-polynomial form with segments defined only by the breakpoints of pp1 and...