Runge Kutta 第二种方法的 Python 代码是什么

Question

这段代码可以吗？

def rKN(x, fx, n, hs):
    k1 = []
    k2 = []
    k3 = []
    k4 = []
    xk = []
    for i in range(n):
        k1.append(fx[i](x)*hs)
    for i in range(n):
        xk.append(x[i] + k1[i]*0.5)
    for i in range(n):
        k2.append(fx[i](xk)*hs)
    for i in range(n):
        xk[i] = x[i] + k2[i]*0.5
    for i in range(n):
        k3.append(fx[i](xk)*hs)
    for i in range(n):
        xk[i] = x[i] + k3[i]
    for i in range(n):
        k4.append(fx[i](xk)*hs)
    for i in range(n):
        x[i] = x[i] + (k1[i] + 2*(k2[i] + k3[i]) + k4[i])/6
    return x

Answer 1

numpy似乎更具可读性：

代码取自 http://www.math-cs.gordon.edu/courses/mat342/python/diffeq.py

def rk2a( f, x0, t ):
    """Second-order Runge-Kutta method to solve x' = f(x,t) with x(t[0]) = x0.
    USAGE:
        x = rk2a(f, x0, t)
    INPUT:
        f     - function of x and t equal to dx/dt.  x may be multivalued,
                in which case it should a list or a NumPy array.  In this
                case f must return a NumPy array with the same dimension
                as x.
        x0    - the initial condition(s).  Specifies the value of x when
                t = t[0].  Can be either a scalar or a list or NumPy array
                if a system of equations is being solved.
        t     - list or NumPy array of t values to compute solution at.
                t[0] is the the initial condition point, and the difference
                h=t[i+1]-t[i] determines the step size h.
    OUTPUT:
        x     - NumPy array containing solution values corresponding to each
                entry in t array.  If a system is being solved, x will be
                an array of arrays.
    NOTES:
        This version is based on the algorithm presented in "Numerical
        Analysis", 6th Edition, by Burden and Faires, Brooks-Cole, 1997.
    """
    n = len( t )
    x = numpy.array( [ x0 ] * n )
    for i in xrange( n - 1 ):
        h = t[i+1] - t[i]
        k1 = h * f( x[i], t[i] ) / 2.0
        x[i+1] = x[i] + h * f( x[i] + k1, t[i] + h / 2.0 )
    return x