给定一组在（X，Y，Z）坐标中定义的点，在任意（X，Y）处插值Z值

Question

Given a set of points in (X, Y, Z) coordinates that are points on a surface, I would like to be able to interpolate Z-values at arbitrary (X, Y) coordinates. I've found some success using mlab.griddata for interpolating values on a grid, but I want to be able to call a general use function for any (X, Y) coordinate.

The set of points form a roughly hemispherical surface. To simplify the problem, I am trying to write a method that interpolates values between known points of the hemisphere defined by the x, y, and z coordinates below. Although there is an analytical solution to find z = f(x, y) for a perfect sphere, such that you don't have to interpolate, the actual set of points will not be a perfect sphere, so we should assume that we need to interpolate values at unknown (X, Y) coordinates. Link to IPython notebook with point data

resolution = 10
u = np.linspace(-np.pi / 2, np.pi / 2, resolution)
v = np.linspace(0, np.pi, resolution)

U, V = np.meshgrid(u, v)

xs = np.sin(U) * np.cos(V) 
ys = np.sin(U) * np.sin(V)
zs = np.cos(U)

I have been using scipy.interpolate.interp2d , which "returns a function whose call method uses spline interpolation to find the value of new points."

def polar(xs, ys, zs, resolution=10):
    rs = np.sqrt(np.multiply(xs, xs) + np.multiply(ys, ys))
    ts = np.arctan2(ys, xs)
    func = interp2d(rs, ts, zs, kind='cubic')
    vectorized = np.vectorize(func)

    # Guesses
    ri = np.linspace(0, rs.max(), resolution)
    ti = np.linspace(0, np.pi * 2, resolution)

    R, T = np.meshgrid(ri, ti)
    Z = vectorized(R, T)
    return R * np.cos(T), R * np.sin(T), Z

Unfortunately I get pretty weird results, similarly to another StackOverflow user who tried to use interp2d .

The most success I have found thus far is using inverse squares to estimate values of Z at (X, Y). But the function is not perfect for estimating values of Z near Z=0.

What can I do to get a function z = f(x, y) given a set of points in (x, y, z)? Am I missing something here... do I need more than a point cloud to reliably estimate a value on a surface?

EDIT:

This is the function that I ended up writing. The function takes input arrays of xs, ys, zs and interpolates at x, y using scipy.interpolate.griddata , which does not require a regular grid. I'm sure there is a smarter way to do this and would appreciate any updates, but it works and I'm not concerned with performance. Including a snippet in case it helps anyone in the future.

def interpolate(x, y, xs, ys, zs):
    r = np.sqrt(x*x + y*y)
    t = np.arctan2(y, x)

    rs = np.sqrt(np.multiply(xs, xs) + np.multiply(ys, ys))
    ts = np.arctan2(ys, xs)

    rs = rs.ravel()
    ts = ts.ravel()
    zs = zs.ravel()

    ts = np.concatenate((ts - np.pi * 2, ts, ts + np.pi * 2))
    rs = np.concatenate((rs, rs, rs))
    zs = np.concatenate((zs, zs, zs))


    Z = scipy.interpolate.griddata((rs, ts), zs, (r, t))
    Z = Z.ravel()
    R, T = np.meshgrid(r, t)
    return Z

Answer 1

You're saying that you've tried using griddata . So why was that not working? griddata also works if the new points are not regularly spaced. For example,

# Definitions of xs, ys and zs
nx, ny = 20, 30
x = np.linspace(0, np.pi, nx)
y = np.linspace(0, 2*np.pi, ny)

X,Y = np.meshgrid(x, y)

xs = X.reshape((nx*ny, 1))
ys = Y.reshape((nx*ny, 1))

## Arbitrary definition of zs
zs = np.cos(3*xs/2.)+np.sin(5*ys/3.)**2

## new points where I want the interpolations
points = np.random.rand(1000, 2)

import scipy.interpolate
zs2 = scipy.interpolate.griddata(np.hstack((xs, ys)), zs, points)

Is this not what you are after?

Answer 2

If I understand your question, you have points xs , ys , zs that are defined by

xs = np.sin(U) * np.cos(V) 
ys = np.sin(U) * np.sin(V)
zs = np.cos(U)

What you want is to be able to interpolate and find a z-value for a given x and y? Why do you need interpolation? The above equations represent a sphere, they can be rewritten as xs*xs + ys*ys + zs*zs = 1 , so there is an easy analytical solution to this problem:

def Z(X, Y):
    return np.sqrt(1-X**2-Y**2)
    ## or return -np.sqrt(1-X**2-Y**2) since this equation has two solutions

unless I misunderstood the question.