我目前正在处理天文数据，其中有彗星图像。由于捕获时间（暮光之城），我想删除这些图像中的背景天空渐变。我为此开发的第一个程序从 Matplotlib 的“ginput”（x，y）中提取用户选择的点，提取每个坐标（z）的数据，然后使用 SciPy 的“griddata”将数据网格化到一个新数组中。
由于假设背景仅略有变化，因此我想将 3d 低阶多项式拟合到这组 (x,y,z) 点。但是，“griddata”不允许输入顺序：
griddata(points,values, (dimension_x,dimension_y), method='nearest/linear/cubic')
关于可以使用的另一个函数或开发 leas-squares 拟合的方法的任何想法，这将允许我控制订单？
Griddata 使用样条拟合。三阶样条曲线与三阶多项式不同（相反，它在每个点上都是不同的三阶多项式）。
如果您只想将 2D、3 阶多项式拟合到您的数据中，请执行以下操作以使用您的所有数据点估计 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()
以下实现polyfit2d使用可用的 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()
根据最小二乘原理，模仿Kington的风格，同时将参数m移到参数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()
如果有人正在寻找拟合特定阶数的多项式（而不是最高幂等于 的多项式order，您可以对接受的答案polyfit和进行此调整polyval：
代替：
ij = itertools.product(range(order+1), range(order+1))
其中，对于order=2给定[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2), (2, 0), (2, 1), (2, 2)]（又名高达 4 次多项式），您可以使用
def xy_powers(order):
powers = itertools.product(range(order + 1), range(order + 1))
return [tup for tup in powers if sum(tup) <= order]
[(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (2, 0)]这返回order=2