我正在嘗試使用 matplotlib 將 plot 散點圖放到 python 中的 3d 連續圖上

Question

我在 3d 圖上繪制了洛倫茲吸引子。 我想通過在洛倫茲圖上繪制代表不同增長率大小的不同顏色的點，來了解培育向量的增長率大小如何影響洛倫茲吸引子。

這是我現在的代碼：

fig = plt.figure(figsize=(8, 8))
ax = fig.add_subplot(111, projection = '3d')

plot 洛倫茲

ax.plot(x1, y1, z1)

添加增長率標記

ax = fig.add_subplot

for k in range(100):
    if (GR[k] <= 0):
      plt.scatter(0.5*k, x1[50*k], c = "y")
    elif (0 < GR[k] <= 3.2):
      plt.scatter(0.5*k, x1[50*k], c = "g")
    elif (3.2 < GR[k] <= 6.4):
      plt.scatter(0.5*k, x1[50*k], c = "b")
    else:
      plt.scatter(0.5*k, x1[50*k], c = "r")

x1、y1、z1組成洛倫茲吸引子，GR為培育向量的增長率，其中前50個為：

  [0.          10.282047    10.8977816    9.94731134   5.09550477
  -2.90817325  -8.55789949 -10.22519406  -7.08646881  -4.03173251
   0.32302345   2.48287221  -0.64753007  -1.22328369   1.14720494
   0.50083297  -1.24334573  -1.97221857   1.48577796   2.20605109
  -1.09659768  -0.82320336   1.23992983   0.32689335  -1.35888724
  -1.8668327    1.79410769   1.84711434  -1.38602027  -0.44126068
   1.28436189   0.27735059  -1.35896733  -1.81959438   1.87149091
   1.53278532  -1.54682835  -0.15104558   1.35899661   0.39353056
  -1.21200428  -1.86788144   1.69061062   1.31533289  -1.6250634
   0.01201846   1.5258175    0.71428205  -0.86708544  -1.95685686]

在plt.scatter(0.5*k, x1[50*k], c = "y") ，出現此錯誤：

TypeError: loop of ufunc does not support argument 0 of type NoneType which has no callable sqrt method

我也嘗試過 plot 這樣的散點圖：

ax = fig.add_subplot
for k in range(100):
    if (GR[k] <= 0):
      plt.scatter(0.5*k, x1[50*k], y1[50*k], z1[50*k], c = "y")
    elif (0 < GR[k] <= 3.2):
      plt.scatter(0.5*k, x1[50*k], y1[50*k], z1[50*k], c = "g")
    elif (3.2 < GR[k] <= 6.4):
      plt.scatter(0.5*k, x1[50*k], y1[50*k], z1[50*k], c = "b")
    else:
      plt.scatter(0.5*k, x1[50*k], y1[50*k], z1[50*k], c = "r")

但是在plt.scatter(0.5*k, x1[50*k], y1[50*k], z1[50*k], c = "y")這行會出現以下錯誤：

TypeError: scatter() got multiple values for argument 'c'

有人能幫忙嗎？

下面是一個應該能夠在本地運行的 MIV：

import numpy as np
from scipy.integrate import solve_ivp
import matplotlib.pyplot as plt

#Initial Conditions
X0 = np.array([1, 1, 1])

#define lorenz function
def lorenz_solveivp(sigma=10, r=28, b=8/3):
    def rhs(t, X):
        return np.array([sigma*(X[1] - X[0]), -X[0]*X[2] + r*X[0] - X[1], X[0]*X[1] - b*X[2]])
    return rhs

#define time and apply solve_ivp
t = np.linspace(0, 50, 5001)  #equispaced points from 0 to 50, timestep of 0.01
rhs_function = lorenz_solveivp()
sol1 = solve_ivp(rhs_function, (0, 50), X0, t_eval=t)

#find the growth rates
def growth_rate(X0, dX, n, dt, ng):
    Xp0 = X0 + dX
    times = np.linspace(0, n*dt, n + 1)
    Xn_save = np.zeros((X0.size, n*(ng-1)+1))
    Xpn_save = np.zeros((Xp0.size, n*(ng-1)+1))
    t = np.zeros((n*(ng - 1) + 1))
    i = 0
    g = np.zeros(ng)

    while i < ng - 1:

        Xn = solve_ivp(lorenz_solveivp(), [0, n*dt], X0, t_eval = times)
        Xpn = solve_ivp(lorenz_solveivp(), [0, n*dt], Xp0, t_eval = times)

        Xn_save[:, n*i: n + 1 + n*i] = Xn.y
        Xpn_save[:, n*i: n + 1 + n*i] = Xpn.y
        t[n*i: n + 1 + n*i] = Xn.t + i*n*dt

        dXb = Xpn.y[:,n] - Xn.y[:,n]

        g[i+1] = np.log(np.linalg.norm(dXb)/np.linalg.norm(dX))/(n*dt)

        P_new = dXb*(np.linalg.norm(dX)/np.linalg.norm(dXb))
        Xp0 = Xn.y[:,n] + P_new
        X0 = Xn.y[:,n]
        i += 1

    return g, Xn_save, t

X0 = np.array([1, 1, 1])
dX = np.array([1, 1, 1])/(np.sqrt(3))
dt = 0.01
n = 8
ng = 1000
GR, Xn_save, t = growth_rate(X0, dX, n, dt, ng)

#plot lorenz function
x1, y1, z1 = sol1.y

fig = plt.figure(figsize=(8, 8)) #specify the size of plot
ax = fig.add_subplot(111, projection = '3d')
ax.plot(x1, y1, z1)

# add growth rate markers
ax = fig.add_subplot
for k in range(100):
    if (GR[k] <= 0):
      plt.scatter(0.5*k, x1[50*k], y1[50*k], z1[50*k], color = "y")
    elif (0 < GR[k] <= 3.2):
      plt.scatter(0.5*k, x1[50*k], y1[50*k], z1[50*k], color = "g")
    elif (3.2 < GR[k] <= 6.4):
      plt.scatter(0.5*k, x1[50*k], y1[50*k], z1[50*k], color = "b")
    else:
      plt.scatter(0.5*k, x1[50*k], y1[50*k], z1[50*k], color = "r")

Answer 1

我修復了你的代碼。 代碼有兩個主要問題

• 默認情況下matplotlib具有二維線圖plt.plot默認為 2d。 我通過直接在你的 3d 軸上繪制來解決這個問題
• 刪除0.5 * k因為我不確定這是在做什么，但我認為它不是 3d 坐標系的一部分。

import numpy as np
from scipy.integrate import solve_ivp
from mpl_toolkits.mplot3d import Axes3D
import matplotlib.pyplot as plt

#Initial Conditions
X0 = np.array([1, 1, 1])

#define lorenz function
def lorenz_solveivp(sigma=10, r=28, b=8/3):
    def rhs(t, X):
        return np.array([sigma*(X[1] - X[0]), -X[0]*X[2] + r*X[0] - X[1], X[0]*X[1] - b*X[2]])
    return rhs

#define time and apply solve_ivp
t = np.linspace(0, 50, 5001)  #equispaced points from 0 to 50, timestep of 0.01
rhs_function = lorenz_solveivp()
sol1 = solve_ivp(rhs_function, (0, 50), X0, t_eval=t)

#find the growth rates
def growth_rate(X0, dX, n, dt, ng):
    Xp0 = X0 + dX
    times = np.linspace(0, n*dt, n + 1)
    Xn_save = np.zeros((X0.size, n*(ng-1)+1))
    Xpn_save = np.zeros((Xp0.size, n*(ng-1)+1))
    t = np.zeros((n*(ng - 1) + 1))
    i = 0
    g = np.zeros(ng)

    while i < ng - 1:

        Xn = solve_ivp(lorenz_solveivp(), [0, n*dt], X0, t_eval = times)
        Xpn = solve_ivp(lorenz_solveivp(), [0, n*dt], Xp0, t_eval = times)

        Xn_save[:, n*i: n + 1 + n*i] = Xn.y
        Xpn_save[:, n*i: n + 1 + n*i] = Xpn.y
        t[n*i: n + 1 + n*i] = Xn.t + i*n*dt

        dXb = Xpn.y[:,n] - Xn.y[:,n]

        g[i+1] = np.log(np.linalg.norm(dXb)/np.linalg.norm(dX))/(n*dt)

        P_new = dXb*(np.linalg.norm(dX)/np.linalg.norm(dXb))
        Xp0 = Xn.y[:,n] + P_new
        X0 = Xn.y[:,n]
        i += 1

    return g, Xn_save, t

X0 = np.array([1, 1, 1])
dX = np.array([1, 1, 1])/(np.sqrt(3))
dt = 0.01
n = 8
ng = 1000
GR, Xn_save, t = growth_rate(X0, dX, n, dt, ng)

#plot lorenz function
x1, y1, z1 = sol1.y

fig = plt.figure(figsize=(8, 8)) #specify the size of plot
ax = fig.add_subplot(111, projection = '3d')
ax.plot(x1, y1, z1)

# add growth rate markers
for k in range(100):
    # simplified expression to set color
    if (GR[k] <= 0):
        c = 'y'
    elif (0 < GR[k] <= 3.2):
        c = 'g'
    elif (3.2 < GR[k] <= 6.4):
        c = 'b'
    else:
        c = 'r'
    ax.scatter(x1[50*k], y1[50*k], z1[50*k], color = c)
fig.show()