How to repeat a function n times

Question

I'm trying to write a function in python that is like:

def repeated(f, n):
    ...

where f is a function that takes one argument and n is a positive integer.

For example if I defined square as:

def square(x):
    return x * x

and I called

repeated(square, 2)(3)

this would square 3, 2 times.

Answer 1

That should do it:

 def repeated(f, n):
     def rfun(p):
         return reduce(lambda x, _: f(x), xrange(n), p)
     return rfun

 def square(x):
     print "square(%d)" % x
     return x * x

 print repeated(square, 5)(3)

output:

 square(3)
 square(9)
 square(81)
 square(6561)
 square(43046721)
 1853020188851841

or lambda-less?

def repeated(f, n):
    def rfun(p):
        acc = p
        for _ in xrange(n):
            acc = f(acc)
        return acc
    return rfun

Answer 2

Using reduce and lamba. Build a tuple starting with your parameter, followed by all functions you want to call:

>>> path = "/a/b/c/d/e/f"
>>> reduce(lambda val,func: func(val), (path,) + (os.path.dirname,) * 3)
"/a/b/c"

Answer 3

Something like this?

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

Answer 4

There is an itertools recipe called repeatfunc that performs this operation.

From itertools recipes:

def repeatfunc(func, times=None, *args):
    """Repeat calls to func with specified arguments.

    Example:  repeatfunc(random.random)
    """
    if times is None:
        return starmap(func, repeat(args))
    return starmap(func, repeat(args, times))

I use a third-party library, more_itertools, that conveniently implements these recipes (optional):

import more_itertools as mit

list(mit.repeatfunc(square, 2, 3))
# [9, 9]

Answer 5

I think you want function composition:

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

def square(x):
  return pow(x, 2)

y = compose(square, 3, 2)
print y

Answer 6

Here's a recipe using reduce:

def power(f, p, myapply = lambda init, g:g(init)):
    ff = (f,)*p # tuple of length p containing only f in each slot
    return lambda x:reduce(myapply, ff, x)

def square(x):
    return x * x

power(square, 2)(3)
#=> 81

I call this power, because this is literally what the power function does, with composition replacing multiplication.

(f,)*p creates a tuple of length p filled with f in every index. If you wanted to get fancy, you would use a generator to generate such a sequence (see itertools) - but note it would have to be created inside the lambda.

myapply is defined in the parameter list so that it is only created once.

Answer 7

Using reduce and itertools.repeat (as Marcin suggested):

from itertools import repeat
from functools import reduce # necessary for python3

def repeated(func, n):
    def apply(x, f):
        return f(x)
    def ret(x):
        return reduce(apply, repeat(func, n), x)
    return ret

You can use it as follows:

>>> repeated(os.path.dirname, 3)('/a/b/c/d/e/f')
'/a/b/c'

>>> repeated(square, 5)(3)
1853020188851841

(after importing os or defining square respectively)