python - 如何绘制由python中的向量给定的结构的表面？

我想在笛卡尔坐标x，y，z中绘制由3D向量给出的数据表面。数据不能用平滑函数表示。

因此，首先我们使用函数eq_points(N_count, r)生成一些伪数据，该函数返回一个数组points，该数组带有对象表面上每个点的x，y，z坐标。数量omega是立体角，现在不重要。

#credit to Markus Deserno from MPI
#https://www.cmu.edu/biolphys/deserno/pdf/sphere_equi.pdf
def eq_points(N_count, r):
    points = []
    a = 4*np.pi*r**2/N_count
    d = np.sqrt(a)
    M_theta = int(np.pi/d)
    d_theta = np.pi/M_theta
    d_phi = a/d_theta
    for m in range(M_theta):
        theta = np.pi*(m+0.5)/M_theta
        M_phi = int(2*np.pi*np.sin(theta)/d_phi)
        for n in range(M_phi):
            phi = 2*np.pi*n/M_phi

            points.append(np.array([r*np.sin(theta)*np.cos(phi),
                                    r*np.sin(theta)*np.sin(phi),
                                    r*np.cos(theta)]))

    omega = 4*np.pi/N_count

    return np.array(points), omega

#starting plotting sequence

fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

points, omega = eq_points(400, 1.)

ax.scatter(points[:,0], points[:,1], points[:,2])
ax.scatter(0., 0., 0., c="r")
ax.set_xlabel(r'$x$ axis')
ax.set_ylabel(r'$y$ axis')
ax.set_zlabel(r'$Z$ axis')

plt.savefig("./sphere.png", format="png", dpi=300)
plt.clf()

结果是下图所示的球体。 蓝点标记了points数组中的数据，而红点是原点。

我想得到这样的东西
取自here。但是，mplot3d教程中的数据始终是平滑函数的结果。除了我用于球面图的ax.scatter()函数之外。

因此，最终我的目标是绘制一些仅显示其表面的数据。通过更改到每个蓝点原点的径向距离来生成此数据。此外，有必要确保每个点都与曲面接触。如何在here上绘制表面，例如在plot_surface()中详细构造？一些实际的实时数据如下所示:

最佳答案

用新规范解决了所有点都接触表面的问题。假设如示例中所示由用户设置角度，则通过计算单位球面上具有相同角度的点形成的船体的单纯形，可以很容易地预先计算构成表面的单纯形的点的索引就像在感兴趣的数据集中一样。然后，我们可以使用这些索引来获取感兴趣的表面。

import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from mpl_toolkits.mplot3d.art3d import Poly3DCollection
from scipy.spatial import ConvexHull

def eq_points(N_count, r):
    points = []
    a = 4*np.pi*r**2/N_count
    d = np.sqrt(a)
    M_theta = int(np.pi/d)
    d_theta = np.pi/M_theta
    d_phi = a/d_theta
    for m in range(M_theta):
        theta = np.pi*(m+0.5)/M_theta
        M_phi = int(2*np.pi*np.sin(theta)/d_phi)
        for n in range(M_phi):
            phi = 2*np.pi*n/M_phi

            points.append(np.array([r*np.sin(theta)*np.cos(phi),
                                    r*np.sin(theta)*np.sin(phi),
                                    r*np.cos(theta)]))

    omega = 4*np.pi/N_count

    return np.array(points), omega

def eq_points_with_random_radius(N_count, r):
    points = []
    a = 4*np.pi*r**2/N_count
    d = np.sqrt(a)
    M_theta = int(np.pi/d)
    d_theta = np.pi/M_theta
    d_phi = a/d_theta
    for m in range(M_theta):
        theta = np.pi*(m+0.5)/M_theta
        M_phi = int(2*np.pi*np.sin(theta)/d_phi)
        for n in range(M_phi):
            phi = 2*np.pi*n/M_phi
            rr = r * np.random.rand()
            points.append(np.array([rr*np.sin(theta)*np.cos(phi),
                                    rr*np.sin(theta)*np.sin(phi),
                                    rr*np.cos(theta)]))

    omega = 4*np.pi/N_count

    return np.array(points), omega


N = 400
pts, _ = eq_points(N, 1.)
pts_rescaled, _ = eq_points_with_random_radius(N, 1.)
extremum = 2.

# plot points
fig = plt.figure()
ax = Axes3D(fig)
ax.scatter(pts_rescaled[:,0], pts_rescaled[:,1], pts_rescaled[:,2])
ax.set_xlim(-extremum, extremum)
ax.set_ylim(-extremum, extremum)
ax.set_zlim(-extremum, extremum)

# get indices of simplices making up the surface using points on unit sphere;
# index into rescaled points
hull = ConvexHull(pts)
vertices = [pts_rescaled[s] for s in hull.simplices]

fig = plt.figure()
ax = Axes3D(fig)
triangles = Poly3DCollection(vertices, edgecolor='k')
ax.add_collection3d(triangles)
ax.set_xlim(-extremum, extremum)
ax.set_ylim(-extremum, extremum)
ax.set_zlim(-extremum, extremum)
plt.show()

关于python - 如何绘制由python中的向量给定的结构的表面？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/47696226/