如何使用args向scipy.integrate.solve_ivp传递数组？

Question

如何将数组传递给 scipy.integate.solve_ivp 函数？现在 u=1.0我想要的是 u=np.random.uniform(-1, 1, 1000).

scipy的版本是1.4.1。

代码是:

import numpy as np
from scipy.integrate import solve_ivp

def func(t, x, u):
    dydt = (-x + u) / 5
    return dydt

y0 = 0
t_span = [0, 10]  
t_eval = np.linspace(0, 10, 1000)
u = 1.0

sol = solve_ivp(func, t_span=t, y0=y0, t_eval=t_eval, args=(u, ))

希望能得到帮助

不要忘记逗号 arg=(u, )，否则你会有一个错误 odepack.error: Extra arguments must be in a tuple. 感谢@熊布朗解决了这个问题，我的代码是： import numpy as np from scipy.integration import odeint def func(t, x, u): dydt = (-x + u) 5 return dydt y0.

Answer 1

我想这可能是可行的。

import numpy as np
from scipy.integrate import solve_ivp

def func(t, x, u):
    dydt = (-x + u(t)) / 5
    return dydt

y0 = 0
t_span = [0, 10]  
t_eval = np.linspace(0, 10, 1000)
u = lambda t: np.random.uniform(-1, 1, 1000)

sol = solve_ivp(func, t_span=t, y0=y0, t_eval=t_eval, args=(u, ))

这里有一个更好的解决方案。谢谢 @Lutz Lehmann

import numpy as np
from scipy.integrate import solve_ivp
from scipy.interpolate import interp1d

def func(t, x, u):
    dydt = (-x + u(t)) / 5
    return dydt

y0 = 0
t_span = [0, 10]  
t_eval = np.linspace(0, 10, 1000)
u_value = np.random.uniform(-1, 1, 1000)
u = interp1d(x=t_eval, y=u_value)

sol = solve_ivp(func, t_span=t, y0=y0, t_eval=t_eval, args=(u, ))