Python 3D多项式曲面拟合，依赖于顺序

Question

我目前正在处理其中有彗星图像的天文数据。由于拍摄时间（黄昏），我想删除这些图像中的背景天空渐变。我开发的第一个程序从Matplotlib的“ginput”（x，y）中取出用户选择的点，为每个坐标（z）提取数据，然后用SciPy的“griddata”将数据网格化为一个新数组。

由于假设背景略有变化，我想将3d低阶多项式拟合到这组（x，y，z）点。但是，“griddata”不允许输入订单：

griddata(points,values, (dimension_x,dimension_y), method='nearest/linear/cubic')

关于可能使用的其他功能的任何想法或开发leas-square拟合的方法，这将允许我控制订单？

Answer 1

Griddata使用样条拟合。三阶样条与三阶多项式不同（相反，它在每个点都是不同的三阶多项式）。

如果您只想将二维，三阶多项式拟合到数据中，请执行以下操作，使用所有数据点估算16个系数。

import itertools
import numpy as np
import matplotlib.pyplot as plt

def main():
    # Generate Data...
    numdata = 100
    x = np.random.random(numdata)
    y = np.random.random(numdata)
    z = x**2 + y**2 + 3*x**3 + y + np.random.random(numdata)

    # Fit a 3rd order, 2d polynomial
    m = polyfit2d(x,y,z)

    # Evaluate it on a grid...
    nx, ny = 20, 20
    xx, yy = np.meshgrid(np.linspace(x.min(), x.max(), nx), 
                         np.linspace(y.min(), y.max(), ny))
    zz = polyval2d(xx, yy, m)

    # Plot
    plt.imshow(zz, extent=(x.min(), y.max(), x.max(), y.min()))
    plt.scatter(x, y, c=z)
    plt.show()

def polyfit2d(x, y, z, order=3):
    ncols = (order + 1)**2
    G = np.zeros((x.size, ncols))
    ij = itertools.product(range(order+1), range(order+1))
    for k, (i,j) in enumerate(ij):
        G[:,k] = x**i * y**j
    m, _, _, _ = np.linalg.lstsq(G, z)
    return m

def polyval2d(x, y, m):
    order = int(np.sqrt(len(m))) - 1
    ij = itertools.product(range(order+1), range(order+1))
    z = np.zeros_like(x)
    for a, (i,j) in zip(m, ij):
        z += a * x**i * y**j
    return z

main()

Answer 2

polyfit2d的以下实现使用了可用的numpy方法numpy.polynomial.polynomial.polyvander2d和numpy.polynomial.polynomial.polyval2d

#!/usr/bin/env python3

import unittest


def polyfit2d(x, y, f, deg):
    from numpy.polynomial import polynomial
    import numpy as np
    x = np.asarray(x)
    y = np.asarray(y)
    f = np.asarray(f)
    deg = np.asarray(deg)
    vander = polynomial.polyvander2d(x, y, deg)
    vander = vander.reshape((-1,vander.shape[-1]))
    f = f.reshape((vander.shape[0],))
    c = np.linalg.lstsq(vander, f)[0]
    return c.reshape(deg+1)

class MyTest(unittest.TestCase):

    def setUp(self):
        return self

    def test_1(self):
        self._test_fit(
            [-1,2,3],
            [ 4,5,6],
            [[1,2,3],[4,5,6],[7,8,9]],
            [2,2])

    def test_2(self):
        self._test_fit(
            [-1,2],
            [ 4,5],
            [[1,2],[4,5]],
            [1,1])

    def test_3(self):
        self._test_fit(
            [-1,2,3],
            [ 4,5],
            [[1,2],[4,5],[7,8]],
            [2,1])

    def test_4(self):
        self._test_fit(
            [-1,2,3],
            [ 4,5],
            [[1,2],[4,5],[0,0]],
            [2,1])

    def test_5(self):
        self._test_fit(
            [-1,2,3],
            [ 4,5],
            [[1,2],[4,5],[0,0]],
            [1,1])

    def _test_fit(self, x, y, c, deg):
        from numpy.polynomial import polynomial
        import numpy as np
        X = np.array(np.meshgrid(x,y))
        f = polynomial.polyval2d(X[0], X[1], c)
        c1 = polyfit2d(X[0], X[1], f, deg)
        np.testing.assert_allclose(c1,
                                np.asarray(c)[:deg[0]+1,:deg[1]+1],
                                atol=1e-12)

unittest.main()

Answer 3

如果有人要拟合特定阶数的多项式（而不是最高幂等于order的多项式，则可以对接受的答案的polyfit和{{1 }}：

代替：

polyval

对于ij = itertools.product(range(order+1), range(order+1)) 给出order=2（又称4阶多项式），您可以使用

[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]

这将返回def xy_powers(order): powers = itertools.product(range(order + 1), range(order + 1)) return [tup for tup in powers if sum(tup) <= order] 的{​​{1}}

Answer 4

根据Least squares的原理，并模仿Kington的风格， 同时将参数m移至参数m_1和参数m_2。

import numpy as np
import matplotlib.pyplot as plt

import itertools


# w = (Phi^T Phi)^{-1} Phi^T t
# where Phi_{k, j + i (m_2 + 1)} = x_k^i y_k^j,
#       t_k = z_k,
#           i = 0, 1, ..., m_1,
#           j = 0, 1, ..., m_2,
#           k = 0, 1, ..., n - 1
def polyfit2d(x, y, z, m_1, m_2):
    # Generate Phi by setting Phi as x^i y^j
    nrows = x.size
    ncols = (m_1 + 1) * (m_2 + 1)
    Phi = np.zeros((nrows, ncols))
    ij = itertools.product(range(m_1 + 1), range(m_2 + 1))
    for h, (i, j) in enumerate(ij):
        Phi[:, h] = x ** i * y ** j
    # Generate t by setting t as Z
    t = z
    # Generate w by solving (Phi^T Phi) w = Phi^T t
    w = np.linalg.solve(Phi.T.dot(Phi), (Phi.T.dot(t)))
    return w


# t' = Phi' w
# where Phi'_{k, j + i (m_2 + 1)} = x'_k^i y'_k^j
#       t'_k = z'_k,
#           i = 0, 1, ..., m_1,
#           j = 0, 1, ..., m_2,
#           k = 0, 1, ..., n' - 1
def polyval2d(x_, y_, w, m_1, m_2):
    # Generate Phi' by setting Phi' as x'^i y'^j
    nrows = x_.size
    ncols = (m_1 + 1) * (m_2 + 1)
    Phi_ = np.zeros((nrows, ncols))
    ij = itertools.product(range(m_1 + 1), range(m_2 + 1))
    for h, (i, j) in enumerate(ij):
        Phi_[:, h] = x_ ** i * y_ ** j
    # Generate t' by setting t' as Phi' w
    t_ = Phi_.dot(w)
    # Generate z_ by setting z_ as t_
    z_ = t_
    return z_


if __name__ == "__main__":
    # Generate x, y, z
    n = 100
    x = np.random.random(n)
    y = np.random.random(n)
    z = x ** 2 + y ** 2 + 3 * x ** 3 + y + np.random.random(n)

    # Generate w
    w = polyfit2d(x, y, z, m_1=3, m_2=2)

    # Generate x', y', z'
    n_ = 1000
    x_, y_ = np.meshgrid(np.linspace(x.min(), x.max(), n_),
                         np.linspace(y.min(), y.max(), n_))
    z_ = np.zeros((n_, n_))
    for i in range(n_):
        z_[i, :] = polyval2d(x_[i, :], y_[i, :], w, m_1=3, m_2=2)

    # Plot
    plt.imshow(z_, extent=(x_.min(), y_.max(), x_.max(), y_.min()))
    plt.scatter(x, y, c=z)
    plt.show()