How can I apply a function to itself?

Question

Suppose I have function, f, which takes in some variable and returns a variable of the same type. For simplicity, let's say

def f(x):
    return x/2+1

I'm interested in applying f to itself over and over. Something like f(f(f(...(f(x))...))).

I could do this like

s = f(x)
for i in range(100):
    s = f(s)

But I was wondering if there was a simpler, less verbose way to doing the same thing. I wan't to avoid for loops (just as a challenge to myself). Is there maybe some way of using map or a similar function to accomplish this?

Answer 1

Is there maybe some way of using map or a similar function to accomplish this?

Not map, but reduce. I wouldn't use it for this, but you could call reduce on an n-item sequence to cause f to be called n times. For example:

>>> def f(x):
...   return x+1
... 
>>> reduce(lambda n,_: f(n), range(100), 42)
142

Explanation:

• n is assigned each successive return value of f.
• _ is the list of numbers from range(100). These numbers are all ignored. All that matters is how many there are.
• 42 is the starting value.

100 nested calls to f(f(f...(f(42))...)) results in 142.

Answer 2

In Python, a for loop is the most ergonomic and readable way to do this. So I would consider this mostly an exercise — these are more natural to use in functional languages.

functools.reduce collapses a list of values to a single value by repeatedly calling a function of two arguments. Here's factorial:

>>> import functools, operator
>>> operator.mul(2,3)
6
>>> functools.reduce(operator.mul, range(1, 10), 1)
362880

We can abuse this to use a list of values for its length only and ignore the actual contents.

>>> def f(x):
...   return x/2+1
... 
>>> functools.reduce(lambda x, y: f(x), range(10), 1)
1.9990234375

Or we can string together n copies of the (unary) function in a list and collapse them by applying each one to the accumulated value.

>>> import itertools
>>> functools.reduce(lambda x, g: g(x), itertools.repeat(f, 10), 1)
1.9990234375