Repeat function python

Question

I'm stuck at higher-order functions in python. I need to write a repeat function repeat that applies the function f n times on a given argument x.

For example, repeat(f, 3, x) is f(f(f(x))).

This is what I have:

def repeat(f,n,x):
    if n==0:
        return f(x)
    else:
        return repeat(f,n-1,x)

When I try to assert the following line:

plus = lambda x,y: repeat(lambda z:z+1,x,y)
assert plus(2,2) == 4

It gives me an AssertionError. I read about How to repeat a function n times but I need to have it done in this way and I can't figure it out...

Is the f(x) function returning anything? Unless otherwise specified, it will return None — Savir
– Savir, Commented Jun 17, 2014 at 12:56
And you want get the result of the last application of f(x)? — thefourtheye
– thefourtheye, Commented Jun 17, 2014 at 12:58
Shouldn't repeat return something like return f(repeat(f,n-1,x))? — tgmath
– tgmath, Commented Jun 17, 2014 at 12:58
As a simpler example, try plus(0,2), which should be 2 but your code gives 3. It doesn't recurse, so it should be easy to debug. — Kevin
– Kevin, Commented Jun 17, 2014 at 12:59
Instead of just asserting, you can try looking at the actual return value of plus. Try it with different inputs and it will be easy to see what the problem is. — interjay
– interjay, Commented Jun 17, 2014 at 13:00

jonrsharpe · Accepted Answer · 2014-06-17 13:01:42Z

5

You have two problems:

You are recursing the wrong number of times (if n == 1, the function should be called once); and
You aren't calling f on the returned value from the recursive call, so the function is only ever applied once.

Try:

def repeat(f, n, x):
    if n == 1: # note 1, not 0
        return f(x)
    else:
        return f(repeat(f, n-1, x)) # call f with returned value

or, alternatively:

def repeat(f, n, x):
    if n == 0:
        return x # note x, not f(x)
    else:
        return f(repeat(f, n-1, x)) # call f with returned value

(thanks to @Kevin for the latter, which supports n == 0).

Example:

>>> repeat(lambda z: z + 1, 2, 2)
4
>>> assert repeat(lambda z: z * 2, 4, 3) == 3 * 2 * 2 * 2 * 2
>>>

answered Jun 17, 2014 at 13:01

jonrsharpe

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1 Comment

Kevin Over a year ago

It may be preferable to do if n == 0: return x rather than if n == 1: return f(x). That way, plus(0,2) will still work.

A_User · Accepted Answer · 2014-06-17 13:06:29Z

2

You've got a very simple error there, in the else block you are just passing x along without doing anything to it. Also you are applying x when n == 0, don't do that.

def repeat(f,n,x):
    """
    >>> repeat(lambda x: x+1, 2, 0)
    2
    """
    return repeat(f, n-1, f(x)) if n > 0 else x

answered Jun 17, 2014 at 13:06

A_User

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