我正在尝试使用 matplotlib 在 python 中将散点图绘制到我的 3d 连续图上
I am trying to plot a scatter graph onto my 3d continuous graph in python using matplotlib
我已经在 3d 图上绘制了洛伦兹吸引子。我想通过在洛伦兹图上绘制代表不同增长率大小的不同颜色的点，来了解培育向量的增长率大小如何影响洛伦兹吸引子。
这是我目前的代码：
fig = plt.figure(figsize=(8, 8))
ax = fig.add_subplot(111, projection = '3d')
情节洛伦兹
ax.plot(x1, y1, z1)
添加增长率标记
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组成洛伦兹吸引子，GR为繁殖向量的增长率，其中前50个为：
[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]
在第 plt.scatter(0.5*k, x1[50*k], c = "y") 行出现此错误：
TypeError: loop of ufunc does not support argument 0 of type NoneType which has no callable sqrt method
我也尝试过这样绘制散点图：
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") 行出现以下错误：
TypeError: scatter() got multiple values for argument 'c'
有人可以帮忙吗？
下面是一个MIV，应该可以在本地运行:
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")
我修复了你的代码。代码有 2 个主要问题
- 默认情况下
matplotlib 有二维线图 plt.plot 默认为二维。我通过直接在你的 3d 轴上绘制来解决这个问题
- 删除了
0.5 * k，因为我不确定这是做什么的，但我认为它不是 3d 坐标系的一部分。
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()
我已经在 3d 图上绘制了洛伦兹吸引子。我想通过在洛伦兹图上绘制代表不同增长率大小的不同颜色的点，来了解培育向量的增长率大小如何影响洛伦兹吸引子。
这是我目前的代码：
fig = plt.figure(figsize=(8, 8))
ax = fig.add_subplot(111, projection = '3d')
情节洛伦兹
ax.plot(x1, y1, z1)
添加增长率标记
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组成洛伦兹吸引子，GR为繁殖向量的增长率，其中前50个为：
[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]
在第 plt.scatter(0.5*k, x1[50*k], c = "y") 行出现此错误：
TypeError: loop of ufunc does not support argument 0 of type NoneType which has no callable sqrt method
我也尝试过这样绘制散点图：
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") 行出现以下错误：
TypeError: scatter() got multiple values for argument 'c'
有人可以帮忙吗？
下面是一个MIV，应该可以在本地运行:
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")
我修复了你的代码。代码有 2 个主要问题
- 默认情况下
matplotlib有二维线图plt.plot默认为二维。我通过直接在你的 3d 轴上绘制来解决这个问题 - 删除了
0.5 * k，因为我不确定这是做什么的，但我认为它不是 3d 坐标系的一部分。
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()