Question

我在三维空间中有一些点，我想通过它们拟合二次曲面。

我试过这段代码

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


def main():
    points = [ [ 175697888, -411724928, 0.429621160030365 ], [ 175697888, -411725144, 0.6078286170959473 ], [ 175698072, -411724640, 0.060898926109075546 ], [ 175698008, -411725360, 0.6184252500534058 ], [ 175698248, -411725720, 0.0771455243229866 ], [ 175698448, -411724456, -0.5925689935684204 ], [ 175698432, -411725936, -0.17584866285324097 ], [ 175698608, -411726152, -0.24736160039901733 ], [ 175698840, -411724360, -1.27967369556427 ], [ 175698800, -411726440, -0.21100902557373047 ], [ 175699016, -411726744, -0.12785470485687256 ], [ 175699280, -411724208, -2.472576856613159 ], [ 175699536, -411726688, -0.19858847558498383 ], [ 175699760, -411724104, -3.5765910148620605 ], [ 175699976, -411726504, -0.7432857155799866 ], [ 175700224, -411723960, -4.770215034484863 ], [ 175700368, -411726304, -1.2959377765655518 ], [ 175700688, -411723760, -6.518451690673828 ], [ 175700848, -411726080, -3.02254056930542 ], [ 175701160, -411723744, -7.941056251525879 ], [ 175701112, -411725896, -3.884831428527832 ], [ 175701448, -411723824, -8.661275863647461 ], [ 175701384, -411725720, -5.21607780456543 ], [ 175701704, -411725496, -6.181706428527832 ], [ 175701800, -411724096, -9.490276336669922 ], [ 175702072, -411724344, -10.066594123840332 ], [ 175702216, -411724560, -10.098011016845703 ], [ 175702256, -411724864, -9.619892120361328 ], [ 175702032, -411725160, -6.936516284942627 ] ]

    n = len(points)
    x, y, z = map(np.array, zip(*points))

    plt.figure()
    plt.subplot(1, 1, 1)

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

    # Evaluate it on a grid...
    nx, ny = 100, 100
    xx, yy = np.meshgrid(np.linspace(x.min(), x.max(), nx), np.linspace(y.min(), y.max(), ny))
    zz = polyval2d(xx, yy, m)

    plt.scatter(xx, yy, c=zz, marker=2)
    plt.scatter(x, y, c=z)
    plt.show()

def polyfit2d(x, y, z, order=2):
    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()

基于这个答案：Python 3D polynomial surface fit, order dependent

但它实际上给出了相反的结果：

查看与表面相比的点的颜色。知道我做错了什么吗？

编辑：更新代码以移除显示不是问题的imshow。

Answer 1

浮点精度似乎存在问题。我稍微使用了您的代码，并更改y和x, y = x - x[0], y - y[0]的范围，使最小二乘解决方案正常工作。做

import itertools
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
# from matplotlib import cbook
from matplotlib import cm
from matplotlib.colors import LightSource


def poly_matrix(x, y, order=2):
    """ generate Matrix use with lstsq """
    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
    return G


points = np.array([[175697888, -411724928, 0.429621160030365],
                   [175697888, -411725144, 0.6078286170959473],
                   [175698072, -411724640, 0.060898926109075546],
                   [175698008, -411725360, 0.6184252500534058],
                   [175698248, -411725720, 0.0771455243229866],
                   [175698448, -411724456, -0.5925689935684204],
                   [175698432, -411725936, -0.17584866285324097],
                   [175698608, -411726152, -0.24736160039901733],
                   [175698840, -411724360, -1.27967369556427],
                   [175698800, -411726440, -0.21100902557373047],
                   [175699016, -411726744, -0.12785470485687256],
                   [175699280, -411724208, -2.472576856613159],
                   [175699536, -411726688, -0.19858847558498383],
                   [175699760, -411724104, -3.5765910148620605],
                   [175699976, -411726504, -0.7432857155799866],
                   [175700224, -411723960, -4.770215034484863],
                   [175700368, -411726304, -1.2959377765655518],
                   [175700688, -411723760, -6.518451690673828],
                   [175700848, -411726080, -3.02254056930542],
                   [175701160, -411723744, -7.941056251525879],
                   [175701112, -411725896, -3.884831428527832],
                   [175701448, -411723824, -8.661275863647461],
                   [175701384, -411725720, -5.21607780456543],
                   [175701704, -411725496, -6.181706428527832],
                   [175701800, -411724096, -9.490276336669922],
                   [175702072, -411724344, -10.066594123840332],
                   [175702216, -411724560, -10.098011016845703],
                   [175702256, -411724864, -9.619892120361328],
                   [175702032, -411725160, -6.936516284942627]])

ordr = 2  # order of polynomial
x, y, z = points.T
x, y = x - x[0], y - y[0]  # this improves accuracy

# make Matrix:
G = poly_matrix(x, y, ordr)
# Solve for np.dot(G, m) = z:
m = np.linalg.lstsq(G, z)[0]


# Evaluate it on a grid...
nx, ny = 30, 30
xx, yy = np.meshgrid(np.linspace(x.min(), x.max(), nx),
                     np.linspace(y.min(), y.max(), ny))
GG = poly_matrix(xx.ravel(), yy.ravel(), ordr)
zz = np.reshape(np.dot(GG, m), xx.shape)

# Plotting (see http://matplotlib.org/examples/mplot3d/custom_shaded_3d_surface.html):
fg, ax = plt.subplots(subplot_kw=dict(projection='3d'))
ls = LightSource(270, 45)
rgb = ls.shade(zz, cmap=cm.gist_earth, vert_exag=0.1, blend_mode='soft')
surf = ax.plot_surface(xx, yy, zz, rstride=1, cstride=1, facecolors=rgb,
                       linewidth=0, antialiased=False, shade=False)
ax.plot3D(x, y, z, "o")

fg.canvas.draw()
plt.show()

解决了准确性问题。你可以尝试：

rank

给出 3DResult Plot]

要评估您的健康质量，请阅读np.linalg.lstsq()的文档。 residual应该是结果向量的大小，$.getJSON('URL.php', function(data) { $.each(data, function(key, val) { ***CODE*** }); }); $.getJSON('URL1.php', function(data) { $.each(data, function(key, val) { ***CODE*** }); }); $.getJSON('URL2.php', function(data) { $.each(data, function(key, val) { ***CODE*** }); });除以数据点数得出平均误差（点与平面之间的距离）。

Answer 2

右侧的表达式

G[:,k] = x**i * y**j

正在无声地溢出。 x和y numpy整数数组，因此结果x**i * y**j也是一个整数数组。但是当i和j都是2时，某些产品会溢出，并回绕到负值。

更改代码以使x和y浮点。例如，

    points = np.array([ [ 175697888, -411724928, 0.429621160030365 ], [ 175697888, -411725144, 0.6078286170959473 ], [ 175698072, -411724640, 0.060898926109075546 ], [ 175698008, -411725360, 0.6184252500534058 ], [ 175698248, -411725720, 0.0771455243229866 ], [ 175698448, -411724456, -0.5925689935684204 ], [ 175698432, -411725936, -0.17584866285324097 ], [ 175698608, -411726152, -0.24736160039901733 ], [ 175698840, -411724360, -1.27967369556427 ], [ 175698800, -411726440, -0.21100902557373047 ], [ 175699016, -411726744, -0.12785470485687256 ], [ 175699280, -411724208, -2.472576856613159 ], [ 175699536, -411726688, -0.19858847558498383 ], [ 175699760, -411724104, -3.5765910148620605 ], [ 175699976, -411726504, -0.7432857155799866 ], [ 175700224, -411723960, -4.770215034484863 ], [ 175700368, -411726304, -1.2959377765655518 ], [ 175700688, -411723760, -6.518451690673828 ], [ 175700848, -411726080, -3.02254056930542 ], [ 175701160, -411723744, -7.941056251525879 ], [ 175701112, -411725896, -3.884831428527832 ], [ 175701448, -411723824, -8.661275863647461 ], [ 175701384, -411725720, -5.21607780456543 ], [ 175701704, -411725496, -6.181706428527832 ], [ 175701800, -411724096, -9.490276336669922 ], [ 175702072, -411724344, -10.066594123840332 ], [ 175702216, -411724560, -10.098011016845703 ], [ 175702256, -411724864, -9.619892120361328 ], [ 175702032, -411725160, -6.936516284942627 ] ])
    x, y, z = points.T

如何将二次曲面拟合到Python中的某些点？

2 个答案: