Question

尊敬的stackoverflow用户，

我是计算机化学家，并且遇到几何问题。我有一堆定义分子表面的坐标，我想导出该表面的向外法向矢量。看起来，当我查看曲面时，曲面近似于流形的属性，尽管并没有使用此框架明确导出坐标点。我还必须明确指出，在一般情况下，分子表面并不总是凸包，而是可以具有一定程度的凹度。他们没有的是不连续，表面通过构造是光滑的。但是由于我不知道如何处理这些数学规范，因此我尝试设计一种解决一般问题的算法。

作为一个初步的评论，我可以确定表面的每个点。因此，对于每个点，我也可以使用这些xyz坐标。该算法采用以下形式：

1。计算每个可用点之间的距离矩阵（不可避免地缩放为点数的平方，但是对于我使用numpy的情况仍然是合理的）

2。。提取每个点的两个最近的邻居

3。。使用此三元组点生成两个以每个点为中心的向量

4。。根据这两个向量的叉积并对其进行归一化，获得法线向量

5。。计算该点与其基础原子之间的向量

6。。如果此向量与法向向量之间的夹角小于90°，则此向量向内指向，因此被其相反的向量代替

此完整过程的结果还可以，但是当我使用matplotlib目视检查结果时，仍然有各种向量与表面平行。这是水分子的matplotlib结果：这是用于比较的水分子表面（可以在其中找到潜在的原子）。忽略表面的颜色编码，它是通过表面电荷进行颜色编码的，这与当前的讨论无关。

此表面是由第三方软件获得的，我无法从该第三方软件获取代码。我只能对其进行可视化，并且无法访问其中用于最终渲染的平滑过程。

如图像所示，表面非常光滑，因此我希望法向矢量能够解决这种“光滑”问题，而这种缺陷并不完美。我需要法线向量来实际反映表面的平滑度，因为当前法线向量所描绘的表面粗糙度对基于这些法线向量的后续计算的质量产生了显着影响。有谁知道我可以在合理的计算时间内解决该问题的方法？

这是一个可以复制我的第一个数字的有效代码：

import math
import matplotlib.pyplot      as plt
import numpy                  as np
from mpl_toolkits.mplot3d     import Axes3D
from sklearn.metrics.pairwise import euclidean_distances


coord = np.array([[ -1.2873729481345813 , 0.03256614731449952 , 1.5416924157851974 ],
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[ 0.764816906209 , 0.0 , -0.200753505843 ],  
[ 0.764816906209 , 0.0 , -0.200753505843 ],  
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[ -1.21073950301e-06 , 0.0 , 0.40150605663 ],
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[ -1.21073950301e-06 , 0.0 , 0.40150605663 ],
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[ -1.21073950301e-06 , 0.0 , 0.40150605663 ],
[ -1.21073950301e-06 , 0.0 , 0.40150605663 ],
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[ 0.764816906209 , 0.0 , -0.200753505843 ],  
[ 0.764816906209 , 0.0 , -0.200753505843 ]])

#transpose coordinate array
xyz = np.transpose(coord)

#establish distance matrix
n = len(coord)
dist_vectors    = list()
dist_vectors.append(xyz[0])
dist_vectors.append(xyz[1])
dist_vectors.append(xyz[2])
dist_vectors    = map(list, zip(*dist_vectors))
d               = euclidean_distances(dist_vectors, dist_vectors)

#find two nearest neighbors for each segment
x1 = np.zeros((n))
x2 = np.zeros((n))
y1 = np.zeros((n))
y2 = np.zeros((n))
z1 = np.zeros((n))
z2 = np.zeros((n))

for i in range(n):
    x_copy = xyz[0]
    y_copy = xyz[1]
    z_copy = xyz[2]

    d1     = np.delete(d[i], i) #removes distance between segment and itself
    x_copy = np.delete(x_copy, i)
    y_copy = np.delete(y_copy, i)
    z_copy = np.delete(z_copy, i)
    j1     = np.argmin(d1)      #get indice of minimum distance
    x1[i]  = x_copy[j1]
    y1[i]  = y_copy[j1]
    z1[i]  = z_copy[j1]
    d2     = np.delete(d1, j1)  #removes minimum distance
    x_copy = np.delete(x_copy, j1)
    y_copy = np.delete(y_copy, j1)
    z_copy = np.delete(z_copy, j1)
    j2     = np.argmin(d2)      #get indice of second minimum distance
    x2[i]  = x_copy[j2]
    y2[i]  = y_copy[j2]
    z2[i]  = z_copy[j2]

#compute normal vector for each segment based on cross product
normal    = list() 
forGraphs = list()
for i in range(n):
    #make vectors for cross product
    v1 = np.zeros((3))
    v1[0] = x1[i] - coord[i][0]
    v1[1] = y1[i] - coord[i][1]
    v1[2] = z1[i] - coord[i][2]
    v2 = np.zeros((3))
    v2[0] = x2[i] - coord[i][0]
    v2[1] = y2[i] - coord[i][1]
    v2[2] = z2[i] - coord[i][2]

    #make cross product and normalize (normal vector should have a unit norm)
    nv = np.cross(v1, v2)

    nv = nv / np.linalg.norm(nv)
    normal.append(nv)

    #check of outwards pointing
    atv    = np.zeros((3))
    atv[0] = atoms[i][0] - coord[i][0]
    atv[1] = atoms[i][1] - coord[i][1]
    atv[2] = atoms[i][2] - coord[i][2]

    th_check = math.acos(np.dot(nv, atv) / (np.linalg.norm(nv) * np.linalg.norm(atv)))
    if th_check < (math.pi / 2): #if inwards pointing (i. e. pointing towards underlying atom), normal vector is replaced by its opposite
        nv[0] = -nv[0]
        nv[1] = -nv[1]
        nv[2] = -nv[2]
    forGraphs.append(np.array([coord[i][0],coord[i][1],coord[i][2],nv[0],nv[1], nv[2]]))

#plot normal vectors (for checkup)
forGraphs = np.asarray(forGraphs)
X, Y, Z, U, V, W = zip(*forGraphs)
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')
ax.quiver(X, Y, Z, U, V, W)
ax.set_xlim([min(xyz[0])- 1, max(xyz[0]) + 1])
ax.set_ylim([min(xyz[1])- 1, max(xyz[1]) + 1])
ax.set_zlim([min(xyz[2])- 1, max(xyz[2]) + 1])
plt.show()

此代码的前280行主要专用于再现结果所必需的坐标表。该代码最重要的部分是从282行到355行，在这里实现了我刚才概述的算法。

在此先感谢您的帮助！

Answer 1

估计法线的“标准”过程是每个点都找到k个最近的邻居，其中k是一个小数（十个？）。然后通过这些点计算最佳平面拟合，并使用该平面的法线。

不幸的是，出现了一个困难，因为正常现象的迹象不确定，您需要实施正常的一致性实施过程。也许就您而言，这很容易，因为所有法线似乎都指向某个中心点。

Answer 2

按照@Yves Daoust的建议，我使用了最近邻技术，并使用奇异值分解（SVD）拟合了通过10个最近邻的飞机。在视觉上容易的情况下，这仍然给出错误的答案。我不明白为什么。这是我得到的样本：

从表面坐标推导流形的法线向量

2 个答案: