我是Python中的初级/中级。我已经编写了一个4th-order Runge-Kutta method (RK4)到Python。它基本上解决了摆锤,但这不是重点。我希望能够将函数f直接传递给RK4函数,即RK4(y_0,n,h)应该成为RK4(f,y_0,n) ,H)。这可以带来很大的好处,我可以将RK4用于描述其他系统的其他f函数,而不仅仅是这一个钟摆。如何将函数传递给Python中的函数?
我曾经玩过简单的功能到RK4,但我做错了什么。我如何在Python中做到这一点?
import numpy as np
def RK4(y_0, n, h):
#4th order Runge-Kutta solver, takes as input
#initial value y_0, the number of steps n and stepsize h
#returns solution vector y and time vector t
#right now function f is defined below
t = np.linspace(0,n*h,n,endpoint = False) #create time vector t
y = np.zeros((n,len(y_0))) #create solution vector y
y[0] = y_0 #assign initial value to first position in y
for i in range(0,n-1):
#compute Runge-Kutta weights k_1 till k_4
k_1 = f(t[i],y[i])
k_2 = f(t[i] + 0.5*h, y[i] + 0.5*h*k_1)
k_3 = f(t[i] + 0.5*h, y[i] + 0.5*h*k_2)
k_4 = f(t[i] + 0.5*h, y[i] + h*k_3)
#compute next y
y[i+1] = y[i] + h/6. * (k_1 + 2.*k_2 + 2.*k_3 + k_4)
return t,y
def f(t,vec):
theta=vec[0]
omega = vec[1]
omegaDot = -np.sin(theta) - omega + np.cos(t)
result = np.array([omega,omegaDot])
return result
test = np.array([0,0.5])
t,y = RK4(test,10,0.1)
感谢您的澄清。我来自MatLab(与一些C)。所以Python不停地惊讶于:-)。 – seb 2013-03-21 10:57:29