我使用SciPy用脚本数字解决了Lorenz方程:
# Lorenz Equations SciPy solver
import numpy as np
from scipy import integrate
from math import cos
from matplotlib import pyplot as plt
a, b = 0, 100
sigma, rho, beta = 10, 28, 8/3
N = 1000000
h = (b-a) / float(N)
def solvr(Y, t):
return [sigma*(Y[1]-Y[0]), Y[0]*(rho-Y[2])-Y[1], Y[0]*Y[1]-beta*Y[2]]
t = np.arange(a, b, h)
asol = integrate.odeint(solvr, [0, 1, 1], t)
x = asol[:,0]
y = asol[:,1]
z = asol[:,2]
现在我想做的是在3D线上相互对抗x
,y
和z
(这些都是Numpy ndarrays,如果你不确定)线框)情节。我认为这必须使用matplotlib完成,但我并不挑剔,只要你给我一个能用3D绘制数据的解决方案我不关心我需要导入哪些模块。
答案 0 :(得分:5)
以下是Lorenz吸引子的3D和动画。该剧本在Jake VanderPlas的以下链接中(以及许多好东西)' Pythonic Perambulations。您可以通过逐行浏览脚本来学到很多东西 - 它是对matplotlib
个对象的优雅使用。
https://jakevdp.github.io/blog/2013/02/16/animating-the-lorentz-system-in-3d/
我在return
函数中animate
之前添加了这两行,然后使用ImageJ导入"图像堆栈"并保存"动画GIF":
fname = "Astro_Jake_" + str(i+10000)[1:]
fig.savefig(fname)
注意:对于OSX,似乎有必要在blit = False
中设置animation.FuncAnimation
。
以下是基于以上内容绘制3D线条的最小简化示例:
def lorentz_deriv((x, y, z), t0, sigma=10., beta=8./3, rho=28.0):
"""Compute the time-derivative of a Lorentz system."""
return [sigma * (y - x), x * (rho - z) - y, x * y - beta * z]
import numpy as np
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d import Axes3D
from scipy.integrate import odeint as ODEint
x = np.linspace(0, 20, 1000)
y, z = 10.*np.cos(x), 10.*np.sin(x) # something simple
fig = plt.figure()
ax = fig.add_subplot(1,2,1,projection='3d')
ax.plot(x, y, z)
# now Lorentz
times = np.linspace(0, 4, 1000)
start_pts = 30. - 15.*np.random.random((20,3)) # 20 random xyz starting values
trajectories = []
for start_pt in start_pts:
trajectory = ODEint(lorentz_deriv, start_pt, times)
trajectories.append(trajectory)
ax = fig.add_subplot(1,2,2,projection='3d')
for trajectory in trajectories:
x, y, z = trajectory.T # transpose and unpack
# x, y, z = zip(*trajectory) # this also works!
ax.plot(x, y, z)
plt.show()
答案 1 :(得分:3)
有关如何在matplotlib网站http://matplotlib.org/mpl_toolkits/mplot3d/tutorial.html#wireframe-plots上进行线框图(以及三维散点图)的简短示例/教程