使用lambda表达式的递归函数 [英] Recursive function using lambda expression

问题描述

我需要创建一个递归函数重复，该重复函数接受一个函数，并使用x的值对n次使用该函数.这是一个迭代版本，可以更详细地说明我的问题.

I need to create a recursive function repeat which takes in a function and uses the function n number of times with the value of x. Here's an iterative version that explains my issue a bit more in detail.

def repeat(fn, n, x):
    res = x
    for i in range(n):
        res = fn(res)
        print(res)
    return res

print(repeat(lambda x: x**2, 3, 3)) returns 6561

首先需要3 ^ 2，然后是3 ^ 2 ^ 2，即81，然后再次是3 ^ 2 ^ 2 ^ 2 = 6561. 我该如何使它递归，使其可以像这样工作.

First it takes 3^2, then 3^2^2 which is 81 then again 3^2^2^2 = 6561. How can i make this recursive so it can work like this.

square_three_times = repeat(lambda x: x**2, 3)
print(square_three_times(3)) return 6561

我尝试过类似的操作，但我真的迷失了方向，不知道该怎么办.

I have tried something like this but im really lost and not sure what to do.

def repeat(fn, n):
    if n == 1:
        return fn(n):
    else:
        def result(x):
            return fn(n)
    return repeat(fn,result(x))

这显然是行不通的，因为递归将永远持续下去.但是我不确定我应该如何编写此代码，因为在进行下一步9 ^ 2之前，我首先需要计算3 ^ 2，依此类推.

This obviously wouldnt work since the recursion would keep going forever. But im not sure how i should write this code since i first need to calculate 3^2 before taking the next step 9^2 and so on.

推荐答案

首先，您弄错了基本情况:

First, you've got the base case wrong:

if n == 1:
    return fn

毕竟，repeat(fn, 1)只是一次应用fn的函数-即fn.

After all, repeat(fn, 1) is just a function that applies fn once—that's fn.

现在，如果基本情况是n == 1，则递归情况几乎总是您将n - 1传递给自己的.

Now, if the base case is when n == 1, the recursive case is almost always going to be something where you pass n - 1 to yourself.

那么，repeat(fn, n)和repeat(fn, n-1)有什么区别?如果您不知道该怎么办，请在您的头上或纸上展开一个简单的盒子:

So, what's the difference between repeat(fn, n) and repeat(fn, n-1)? If you can't figure it out, expand a simple case out in your head or on paper:

repeat(fn, 3)(x): fn(fn(fn(x)))
repeat(fn, 2)(x): fn(fn(x))

现在很明显:repeat(fn, n)与fn(repeat(fn, n-1))是同一回事，对吗?所以:

And now it's obvious: repeat(fn, n) is the same thing as fn(repeat(fn, n-1)), right? So:

else:
    def new_fn(x):
        return fn(repeat(fn, n-1)(x))
    return new_fn

---

但是，正如电影制片人在评论中指出的那样，在此处使用partial会更容易:

---

However, as filmor points out in the comments, it would be easier to use partial here:

def repeat3(fn, n, x):
    if n == 1:
        return fn(x)
    else:
        return fn(repeat3(fn, n-1, x))

def repeat(fn, n):
    return functools.partial(repeat3, fn, n)

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