如何连接 3D 空间中的散点? [英] How can I connect scatter points in a 3D space?
问题描述
我想在 matplotlib 中将所有弧的起点和终点连接在一起.红色弧线是周围所有四个绿色弧线的中心.您将在下面找到绘制平行弧线的代码.
I want to connect the starting point and ending point of all arcs together in matplotlib. Red arc is the centre of all four surrounding green arcs. Below you'll find a code to draw parallel arcs.
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt
import numpy as np
from matplotlib import interactive
interactive(True)
def cart2sphere(x, y, z):
r = np.sqrt(x**2 + y**2 + z**2)
theta = np.arccos(z, r)
phi = np.arctan2(y, x)
return(r, theta, phi)
def sphere2cart(r, theta, phi):
theta = theta - np.pi/2
x = r * np.sin(theta)* np.cos(phi)
y = r * np.sin(theta)* np.sin(phi)
z = r * np.cos(theta)
return(x, y, z)
def pol2cart(rho, phi):
x = rho * np.cos(phi)
y = rho * np.sin(phi)
return(x, y)
# define values
theta = np.pi/2 # arclength in radians
radius = 10 # raduis of circle
k = 1/radius # if you want to use k instead of radius
phi = np.pi/6 # angle of circle in xy plane
radius1 = 7.5
radius2 = 12.5
# discretize for plotting
arcIndex = np.linspace(0, theta, num = 100)
X, Y, Z, = sphere2cart(radius, arcIndex, phi)
X1, Y1, Z1, = sphere2cart(radius2, arcIndex, phi)
X2, Y2, Z2, = sphere2cart(radius1, arcIndex, phi)
X3, Y3, Z3, = sphere2cart(radius1, arcIndex, phi)
X4, Y4, Z4, = sphere2cart(radius2, arcIndex, phi)
# move center or arc to xy plane
# =============================================================================
x1, y1 = pol2cart(radius, phi) #arc
X += x1
Y += y1
x2, y2 = pol2cart(radius, phi) #L1
X1 += x2
Y1 += 1.5+y2
x3, y3 = pol2cart(radius, phi) #L2
X2 += x3
Y2 += y3-1.5
x4, y4 = pol2cart(radius, phi) #L3
X3 += x4
Y3 += 1.5+y4
x5, y5 = pol2cart(radius, phi) #L4
X4 += x5
Y4 += y5-1.5
# =============================================================================
fig = plt.figure()
ax = fig.gca(projection='3d')
# plot arc
ax.plot(X, Y, Z, c= "red", label='arc')
ax.plot(X1, Y1, Z1, c= "Green", label='L1')
ax.plot(X2, Y2, Z2, c= "Green", label='L2')
ax.plot(X3, Y3, Z3, c= "Green", label='L3')
ax.plot(X4, Y4, Z4, c= "Green", label='L4')
# plot axes
ax.plot(np.zeros(100), np.zeros(100), np.linspace(-np.max(np.abs(Z)), np.max(np.abs(Z)), 100), c= "black", alpha = 0.15)
ax.plot(np.zeros(100), np.linspace(-np.max(np.abs(Z)), np.max(np.abs(Z)), 100), np.zeros(100), c= "black", alpha = 0.15)
ax.plot(np.linspace(-np.max(np.abs(Z)), np.max(np.abs(Z)), 100), np.zeros(100), np.zeros(100), c= "black", alpha = 0.15)
# plot center of circle
ax.scatter(np.array([x1]), np.array([y1]), np.array([0]), c = 'Green', label = "center of circle", s=10)
# plot endpoint
# =============================================================================
ax.scatter(X[0], Y[0], Z[0], color="red", label= "origin", s=30)
ax.scatter(X[-1], Y[-1], Z[-1], color= 'red', label = "endpoint", s=30)
ax.scatter(X1[-1], Y1[-1], Z1[-1], color= 'red', s=30)
ax.scatter(X2[-1], Y2[-1], Z2[-1], color= 'red', s=30)
ax.scatter(X3[-1], Y3[-1], Z3[-1], color= 'red', s=30)
ax.scatter(X4[-1], Y4[-1], Z4[-1], color= 'red', s=30)
ax.scatter(X1[1], Y1[1], Z1[1], color= 'red', s=30)
ax.scatter(X2[1], Y2[1], Z2[1], color= 'red', s=30)
ax.scatter(X3[1], Y3[1], Z3[1], color= 'red', s=30)
ax.scatter(X4[1], Y4[1], Z4[1], color= 'red', s=30)
# =============================================================================
# plot projection on each axis
ax.plot(X, np.zeros(len(X)), np.zeros(len(X)), color = "blue", label = "X projection")
ax.plot(np.zeros(len(X)), Y, np.zeros(len(X)), color = "green", label = "Y projection")
ax.plot(np.zeros(len(X)), np.zeros(len(X)), Z, color = "brown", label = "Z projection")
ax.legend()
我从下面显示的代码中得到的结果，
Result that I got from this code shown below,
我正在尝试将所有这些红色散点连接在一起，以便它们可以在弧的起点和终点形成一个正方形.我想要的是如下图所示的东西，
I'm trying to connect all these red scatter points together such that they can make a square at the starting and ending of the arc. What I want is something just like the figure shown below,
如何将空间中的每个点与下一个连接起来，使其成为平滑的折线图?
How do I connect each point in the space with the next one to make it a smooth line graph?
推荐答案
我建议根本不要使用 Matplotlib 进行 3d 绘图.你迟早会撞墙.即使周围有看起来很酷的图，它也不能很好地工作，特别是如果您想将具有许多数据点的散点图可视化.如果您想绘制彩色三角形网格或复杂曲面，它肯定会失败.
I recommend to not use Matplotlib for 3d plotting at all. You will hit a wall sooner or later. Even though there are cool looking plots around, it does not work well especially if you want to visualize scatter plots with many data points. And it will fail for sure if you want to plot colorized triangular meshes or complex surfaces.
改为使用以下工具包之一:
Instead use one of the following toolkits:
- https://docs.enthought.com/mayavi/mayavi/ (3d，快，好)
http://www.pyqtgraph.org/(2d/3d，非常快)
- https://docs.enthought.com/mayavi/mayavi/ (3d, fast, nice)
http://www.pyqtgraph.org/ (2d/3d, very fast)
可以找到一个很好的例子，展示了多个实时绘图的不同方法这里
A very good example showing different approaches for multiple real time plots can be found here
https://plot.ly/python/3d-surface-plots/(2d/3d，不错)
http://vispy.org/documentation.html(2d/3d，非常快，不错，需要 OpenGL)
http://vispy.org/documentation.html (2d/3d, very fast, nice, requires OpenGL)
可能是未来，检查高级函数 vispy.plot() 和
Might be the future, check the high-level functions vispy.plot() and vispy.scene(), I guess they made some progress)
有关示例，请参阅此处，查找带有场景".似乎也有一些接口可以使用 Vispy 来渲染Matplotlib 绘图.
For examples see here, look for the ones with "scene". There also seems to be some interface to use Vispy to render Matplotlib plots.
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