这基本上和OP写的答案是一样的，但我认为它可能更干净一点。首先，我使用了flatten方法，它不像np.ndarray.flatten函数那样冗长。其次，我将结果展平以传递给ODE函数，在函数内部进行整形以提取值并使用，然后将结果展平以返回。最后，我使用np.split来提取每个解。

import numpy as np
from scipy.integrate import odeint

def pend(y, t, b, c, N):
    theta, omega = y.reshape(-1,N)
    dydt = np.stack([omega, -b*omega - c*np.sin(theta)])
    return dydt.flatten()

b = 0.25
c = 5.0
t = np.linspace(0, 10, 101)

# Number of inputs
N = 3

# 1D initial values, directly from docs
theta0 = np.ones((N,)) * (np.pi - 0.1)
omega0 = np.zeros((N,))
y0_2 = np.stack([theta0, omega0])

#   Flattened version of the inputs
y0_2f = y0_2.flatten()

# Run altered pend (pend2) with flattened inputs
sol_2 = odeint(pend, y0_2f, t, args=(b, c, N))
sols_split = np.split(sol_2, N, axis=1)

根据hpaulj的评论，我想出了一个解决方案。这不是很好，因为我传递了N，但它确实有效，而且几乎肯定比循环更好。这应该也适用于N维输入，它只需要在最后重新塑造回原始形状。

import numpy as np
from scipy.integrate import odeint

def pend(y, t, b, c):
    theta, omega = y
    dydt = [omega, -b*omega - c*np.sin(theta)]
    return dydt

def pend2(y, t, b, c, N):
    theta = y[:N]
    omega = y[N:]
    dydt = [omega, -b*omega - c*np.sin(theta)]
    dydt = np.ndarray.flatten(np.array(dydt))
    return dydt

b = 0.25
c = 5.0
t = np.linspace(0, 10, 101)

#   0D initial values, directly from docs
y0_1 = [np.pi - 0.1, 0]
sol_1 = odeint(pend, y0_1, t, args=(b, c))

#   1D initial values, directly from docs
theta0 = np.ones((3)) * (np.pi - 0.1)
omega0 = np.zeros((3))
y0_2 = [theta0, omega0]

#   Flattened version of the inputs
y0_2f = np.ndarray.flatten(np.array(y0_2))

#   Number of inputs
N = len(theta0)

#   Run altered pend (pend2) with flattened inputs
sol_2 = odeint(pend2, y0_2f, t, args=(b, c, N))