python - 递归计算迭代次数

Question

我试图通过这个递归来跟踪迭代，但“iter”总是返回 1。

def compute_root(poly, x_0, epsilon): ##Using Newton's method
    iter = 0
    try:
        if abs(evalpoly(poly,x_0)) < epsilon:
            return '%s is the root. Took %s iterations.'%(x_0,iter)
        else:
            iter += 1
            x_1 = x_0 - (evalpoly(poly,x_0)/evalderiv(poly,x_0))
            return compute_root(poly, x_1, epsilon)

    except ZeroDivisionError:
        print 'No root.'

我想这是因为我必须实际返回 iter += 1 但我不确定。 . .对不起，如果这是一个基本问题。谢谢!

Answer 1

您应该将 iter 传递给您正在递归调用的函数:

def compute_root(poly, x_0, epsilon, iter=None): ##Using Newton's method
    if iter is None:
        iter = 0
    try:
        if abs(evalpoly(poly,x_0)) < epsilon:
            return '%s is the root. Took %s iterations.'%(x_0,iter)
        else:
            iter += 1
            x_1 = x_0 - (evalpoly(poly,x_0)/evalderiv(poly,x_0))
            return compute_root(poly, x_1, epsilon, iter=iter)

    except ZeroDivisionError:
        print 'No root.'