Repeated Function Application

I'm having trouble with a question which follows: Write a recursive function repeatedlyApply that takes a开发者_如何学Pythons arguments a function f of one argument and a positive integer n. The result of repeatedlyApply is a function of one argument that applies f to that argument n times.

So, for example, we would have

repeatedlyApply(lambda x: x+1,10)(100) ==> 110

You may assume that the following function has been defined. You don't have to use it, but it can contribute to a pretty solution.

def compose(f,g):
    return lambda x: f(g(x))

So far i've written this

def compose(f,g):
    return lambda x: f(g(x))

def recApply(f,n):
    for i in range(n):
        return recApply(compose(f,f), n-1)
    return f

I'm going wrong somewhere because using the above example recApply(lambda x: x+1,10)(100) i get 1124.

Help much appreciated

---

Correct answer is:

def recApply(func, n):
    if n > 1:
        rec_func = recApply(func, n - 1)
        return lambda x: func(rec_func(x))
    return func

And the output:

>>>> print recApply(lambda x: x+1,10)(100)
110

---

Your function needs some work:

• You have a return inside your for loop, so you return immediately instead of running the loop.
• You have a recursive call inside your for loop, so you are doing a bit too much iteration. Choose one or the other.
• Be careful when you stack function compositions on top of each other, you are doing power composition rather than linear composition.

Can you tell us what precisely you are trying to do?

EDIT: Since everybody else is posting an answer:

recApply = lambda f, n: lambda x: x if n == 0 else recApply(f, n-1)(f(x))

---

I have a solution based on lambdas:

>>> f = lambda x: x + 10
>>> iterate = lambda f, n, x : reduce(lambda x, y: f(x), range(n), x)
>>> iterate(f, 10, 3)
103
>>> iterate(f, 4, 4)
44
>>> f10 = lambda x: iterate(f, 10, x)
>>> f10(5)
105

---

I assume this is an exercise of some sort. There are a few ways you could do it, here's a short one:

>>> repeatedlyApply = lambda f, n: reduce(lambda f1, f2: compose(f1, f2), [f]*n)
>>> repeatedlyApply(lambda x: x+1,10)(100)
110