我正在尝试绘制简单摆的动画 使用模型https://matplotlib.org/gallery/animation/double_pendulum_sgskip.html。
我的代码如下:
from numpy import sin, cos
import numpy as np
import matplotlib.pyplot as plt
import scipy.integrate as integrate
import matplotlib.animation as animation
#some constants
g = 9.81
l = 0.1
m = 0.01
def sh(r,t):
theta = r[0]
omega = r[1]
sh_theta = omega
sh_omega = -g/l*sin(theta)
return np.array([sh_theta,sh_omega],float)
init_state = [np.radians(89.0),0]
time = np.arange(0,50.0,0.025)
time_elapsed = time
def step_solver(eq, ist, dt):
"""
Execute one time step of length dt and update status
"""
global time_elapsed
state1,state2 = integrate.odeint(eq,ist,[0,dt])
time_elapsed += dt
return state1, state2,time_elapsed
dt = 1/30
ysol,ysolv,timex = step_solver(sh,init_state,dt)
print("This is the y0 values: ", ysol,"y values",ysolv,"This is the time ", timex)
##===================================
##Setting up figure and animation
#======================================
fig = plt.figure()
ax = plt.axes(xlim = (0,2), ylim = (-2,2))
line, = ax.plot([],[],lw=2)
#time_text = ax.text(0.02,0.95,'',transform = ax.transAxes)
#==========================================
##initialisation function: plot the background of each frame
def init():
line.set_data([],[])
#time_text.set_text('')
return line,
def animate(ysol,timex):
x = timex
y = ysol
line.set_data(x,y)
#time_text.set_text(str(i))
return line,
#================================================
#Call the animator
#================================================
anim = animation.FuncAnimation(fig,animate(ysolv,timex), init_func = init, frames = 200, interval =20, blit = True)
anim.save('basic_animation.mp4', fps=30, extra_args=['-vcodec', 'libx264'])
plt.show()
编辑: 我收到错误消息:
ValueError: shape mismatch: objects cannot be broadcast to a single shape
<Figure size 432x288 with 1 Axes>
我已经检查了我的函数step_solver用来打印什么
print(len(ysol),len(ysolv),len(timex))
,在推荐之后,另一个odeint的y输出给出了我的ysol和ysolv变量仅2个值,共2200个时间值。
我期望每个时间步长都能得到y值范围。 我不确定如何排序。我的函数step_solver是否编码错误? 为什么我只得到2个值,如何以与双摆相同的方式为解决方案制作动画?
关于问题可能出在哪里的任何建议? 提前谢谢了。
答案 0 :(得分:0)
让我们更紧密地遵循示例中的代码。这意味着只使用一个积分调用,然后在动画中使用该调用结果的切片。
from numpy import sin, cos
import numpy as np
import matplotlib.pyplot as plt
import scipy.integrate as integrate
import matplotlib.animation as animation
#some constants
g = 9.81
l = 0.1
m = 0.01
def sh(r,t):
theta, omega = r
sh_theta = omega
sh_omega = -g/l*sin(theta)
return np.array([sh_theta,sh_omega],float)
init_state = np.radians([89.0,0])
dt = 1.0/30
time = np.arange(0,50.0,dt)
state = integrate.odeint(sh,init_state,time)
##===================================
##Setting up figure and animation
#======================================
fig = plt.figure()
ax = plt.axes(xlim = (0,10), ylim = (-2,2))
line, = ax.plot([],[],lw=2)
#==========================================
##initialisation function: plot the background of each frame
def init():
return line,
def animate(i):
x = time[i:i+30]
y = state[i:i+30,0]
line.set_data(x,y)
return line,
#================================================
#Call the animator
#================================================
anim = animation.FuncAnimation(fig,animate, init_func = init, frames = 200, interval =20, blit = True)
plt.show()
对于使用yield
的步进发生器的变体,请参见Error in RK4 algorithm in Python中的答案。这允许为集成循环封装数据,而无需为其定义类。但是,即使具有动画效果,带有两个样本的函数图如何帮助仍不清楚。
要对摆锤本身进行动画处理,请使用
##===================================
##Setting up figure and animation
#======================================
fig = plt.figure(figsize=(8,6))
ax = plt.axes(xlim = (-2*l,2*l), ylim = (-2*l,l))
line, = ax.plot([],[],'-o',lw=2,ms=8)
#==========================================
##initialisation function: plot the background of each frame
def init():
return line,
def animate(i):
phi = state[i,0]
line.set_data([0,l*sin(phi)],[0,-l*cos(phi)])
return line,
#================================================
#Call the animator
#================================================
anim = animation.FuncAnimation(fig,animate, init_func = init, frames = len(time), interval =100, blit = True)
plt.show()