Craig Baker - 1 year ago 380

Python Question

I have a set of points which approximate a 2D curve. I would like to use Python with numpy and scipy to find a cubic Bézier path which approximately fits the points, where I specify the exact coordinates of two endpoints, and it returns the coordinates of the other two control points.

I initially thought

`scipy.interpolate.splprep()`

My question is similar to this one: How can I fit a Bézier curve to a set of data?, except that they said they didn't want to use numpy. My preference would be to find what I need already implemented somewhere in scipy or numpy. Otherwise, I plan to implement the algorithm linked from one of the answers to that question, using numpy: An algorithm for automatically fitting digitized curves.

Thank you for any suggestions!

Edit: I understand that a cubic Bézier curve is not guaranteed to pass through all the points; I want one which passes through two given endpoints, and which is as close as possible to the specified interior points.

Answer

Here is a piece of python code for fitting points:

```
'''least square qbezier fit using penrose pseudoinverse
>>> V=array
>>> E, W, N, S = V((1,0)), V((-1,0)), V((0,1)), V((0,-1))
>>> cw = 100
>>> ch = 300
>>> cpb = V((0, 0))
>>> cpe = V((cw, 0))
>>> xys=[cpb,cpb+ch*N+E*cw/8,cpe+ch*N+E*cw/8, cpe]
>>>
>>> ts = V(range(11))/10
>>> M = bezierM (ts)
>>> points = M*xys #produces the points on the bezier curve at t in ts
>>>
>>> control_points=lsqfit(points, M)
>>> linalg.norm(control_points-xys)<10e-5
True
>>> control_points.tolist()[1]
[12.500000000000037, 300.00000000000017]
'''
from numpy import array, linalg, matrix
from scipy.misc import comb as nOk
Mtk = lambda n, t, k: t**(k)*(1-t)**(n-k)*nOk(n,k)
bezierM = lambda ts: matrix([[Mtk(3,t,k) for k in range(4)] for t in ts])
def lsqfit(points,M):
M_ = linalg.pinv(M)
return M_ * points
```

Generally on bezier curves check out Animated bezier and bezierinfo