三角剖分 Plot python 弯曲散点数据

Question

我正在尝试从 matplotlib 获取带三角测量的插值轮廓曲面。我的数据看起来像一条曲线，我无法摆脱曲线下方的数据。 我想将外部数据点作为边界。

我从本教程中获得了代码

import matplotlib.tri as tri
fig, (ax1, ax2) = plt.subplots(nrows=2)
xi = np.linspace(-10,150,2000)
yi = np.linspace(-10,60,2000)
triang = tri.Triangulation(x_after, y_after)

interpolator = tri.LinearTriInterpolator(triang, strain_after)
Xi, Yi = np.meshgrid(xi, yi)
zi = interpolator(Xi, Yi)

ax1.triplot(triang, 'ro-', lw=5)
ax1.contour(xi, yi, zi, levels=30, linewidths=0.5, colors='k')
cntr1 = ax1.contourf(xi, yi, zi, levels=30, cmap="jet")
fig.colorbar(cntr1, ax=ax1)
ax1.plot(x_after, y_after, 'ko', ms=3)

ax2.tricontour(x_after, y_after, strain_after, levels=30, linewidths=0.5, colors='k')
cntr2 = ax2.tricontourf(x_after, y_after, strain_after, levels=30, cmap="jet")
fig.colorbar(cntr2, ax=ax2)
ax2.plot(x_after, y_after, 'ko', ms=3)

plt.subplots_adjust(hspace=0.5)
plt.show()

我找到了使用此代码屏蔽数据的选项，但我不知道如何定义屏蔽以获得我想要的内容

triang.set_mask()

这些是内部曲线的值：

    x_after y_after z_after strain_after
39  117.2757    8.7586  0.1904  7.164
40  119.9474    7.152   0.1862  6.6456
37  111.8319    12.0568 0.1671  6.273
38  114.5314    10.4186 0.1651  5.7309
41  122.7482    5.4811  0.1617  9.1563
36  108.8823    13.4417 0.1421  8.8683
42  125.5035    3.8309  0.141   9.7385
33  99.8064 17.6315 0.1357  9.8613
32  96.8869 18.6449 0.1197  4.4147
35  105.8846    14.6086 0.1079  7.7055
28  84.2221 22.0191 0.1076  6.2098
26  77.8689 23.158  0.1067  7.5833
29  87.354  21.2974 0.1044  11.4365
27  81.0778 22.6443 0.1019  8.3794
24  71.4004 23.7749 0.0968  8.6207
34  102.8772    15.9558 0.0959  18.2025
23  68.2124 23.962  0.0939  7.9201
25  74.6905 23.4465 0.0901  9.0361
30  90.5282 20.398  0.0864  14.1051
31  93.802  19.335  0.0794  10.4563
43  128.3489    2.1002  0.0689  9.0292
22  65.0282 24.1107 0.0654  7.99
21  61.7857 24.0129 0.0543  8.2589
20  58.5831 23.9527 0.0407  9.0087
0   -0.0498 -0.5159 0.0308  7.1073
19  55.3115 23.7794 0.0251  9.6441
5   12.5674 9.3369  0.0203  7.2051
2   4.8147  3.6074  0.0191  8.0103
1   2.363   1.5329  0.0184  7.8285
18  52.0701 23.526  0.016   8.0149
3   7.4067  5.5988  0.0111  8.9994
7   18.2495 12.5836 0.0098  9.771
9   23.9992 15.4145 0.0098  6.7995
16  45.5954 22.5274 0.0098  12.9428
4   9.9776  7.5563  0.0093  6.9804
17  48.9177 23.0669 0.0084  9.3782
13  35.9812 20.0588 0.0066  9.6005
6   15.3578 11.0027 0.0062  9.7801
15  42.2909 21.8663 0.0052  12.0288
11  29.816  17.8723 0.0049  8.9085
8   21.1241 14.0893 0.0032  6.5716
10  26.8691 16.7093 0.0014  6.9672
44  131.1371    0.4155  0.0 11.9578
14  39.0687 20.991  -0.0008 9.9907
12  32.9645 18.9796 -0.0102 9.3389
45  134.083 -1.3928 -0.0616 15.29

Answer 1

我的大部分代码都用于构建x 、 y arrays 和节点编号方面的三角形列表，但我想您已经拥有（或者至少您可以获得）网格程序中的三角形列表你用过的……如果你有三角形，它就像plt.tricontourf(x, y, triangles, z)一样简单。

这是无聊内容的完整代码。

import matplotlib.pyplot as plt
import numpy as np

th0 = th2 = np.linspace(-0.5, 0.5, 21)
th1 = np.linspace(-0.475, 0.475, 20)
r = np.array((30, 31, 32))

x = np.concatenate(( np.sin(th0)*r[0], 
                     np.sin(th1)*r[1],
                     np.sin(th2)*r[2]))
y = np.concatenate(( np.cos(th0)*r[0], 
                     np.cos(th1)*r[1],
                     np.cos(th2)*r[2]))

z = np.sin(x)-np.cos(y)

nodes0 = np.arange( 0, 21, dtype=int)
nodes1 = np.arange(21, 41, dtype=int)
nodes2 = np.arange(41, 62, dtype=int)

triangles = np.vstack((
    np.vstack((nodes0[:-1],nodes0[1:],nodes1)).T,
    np.vstack((nodes0[1:-1],nodes1[1:],nodes1[:-1])).T,
    np.vstack((nodes2[:-1],nodes1,nodes2[1:])).T,
    np.vstack((nodes1[:-1],nodes1[1:],nodes2[1:-1])).T,
    (0, 21, 41),
    (20, 61, 40)
    ))

fig, ax = plt.subplots()
ax.set_aspect(1)
tp = ax.triplot(x, y, triangles, color='k', lw=0.5, zorder=4)
tc = ax.tricontourf(x, y, triangles, np.sin(x)-np.cos(y))
plt.colorbar(tc, location='bottom')
plt.show()

Answer 2

我设法通过使用以下代码找到一种方法来避免底部的三角形 plot：

xtri = x_after[triangles] - np.roll(x_after[triangles], 1, axis=1)
ytri = y_after[triangles] - np.roll(y_after[triangles], 1, axis=1)
maxi = np.max(np.sqrt(xtri**2 + ytri**2), axis=1)
max_radius = 4.5
triang.set_mask(maxi > max_radius)