List of functions->single function (in python)

Question

Say that I have a list of functions: [f1, f2, f3] (in python)

How do I return a single function F:=f3(f2(f1())) (notice that F is of function type). I know that it's possible to do it with .reduce() but I was wondering if there's another way to do it without using libraries.

edit:

Also, notice that the length of the list can be greater than 3

I tried:

def func(list):
  i = 1
  new_function = filters[0]
  while i<=len(filters)-1:
      new_function = filters[i](new_function)
      i+=1
  return new_function

but it doesn't work

Where did you start with this problem? Did you manage to write any code for it? — quamrana
– quamrana, Commented Sep 18, 2021 at 20:46
reduce is part of the Python Standard Library and has a rough equivalent available for use in the docs. — Henry Ecker
– Henry Ecker ♦, Commented Sep 18, 2021 at 20:46
def func(List): i = 1 new_function = filters[0] while i<=len(filters)-1: new_function = filters[i](new_function) i+=1 return new_function I tried this by it doesn't work — Onini
– Onini, Commented Sep 18, 2021 at 20:47

trincot · Accepted Answer · 2021-09-18 21:11:56Z

3

The problem in your code is that you pass a function as argument with filters[i](new_function).

I would suggest this recursive solution:

def chain(first, *rest):
    return lambda x: first(chain(*rest)(x) if rest else x)

Example use:

def inc(x):
    return x + 1

def double(x):
    return x * 2

def square(x):
    return x * x

f = chain(square, double, inc)

print(f(5))  # ((5+1)*2) ** 2 == 144

answered Sep 18, 2021 at 21:11

trincot

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Comments

quamrana · Accepted Answer · 2021-09-18 20:52:23Z

0

I see that in the code you tried, you never actually call the first of your functions. (I also assume that your code starts: def func(filters):

Taking into account that f1() takes no parameter, but the rest take the parameter of the return of the previous function, this should work:

def fit(funcs):
    v = funcs[0]()
    for f in funcs[1:]:
        v = f(v)
    return v

def f1():
    return 42

def f2(x):
    return x

def f3(x):
    return x

fs = [f1, f2, f3]

a = lambda:fit(fs)

print(a())

Output: 42

answered Sep 18, 2021 at 20:52

quamrana

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Comments

Nir H. · Accepted Answer · 2021-09-18 21:02:00Z

0

def get_single_func(func_list):
    def single_func(*args, **kwargs):
        ret = func_list[0](*args, **kwargs)
        for func in func_list[1:]:
            ret = func(ret)
        return ret
    return single_func

answered Sep 18, 2021 at 21:02

Nir H.

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2 Comments

Onini Over a year ago

OK after thinking about the problem for a little bit I came up with my own solution. I'm still quite interested in this one. can you explain what (*args, **kwargs) is doing in func_list[0](*args, **kwargs)?

Nir H. Over a year ago

The operators * and ** are called "unpacking operators". The first one converts a list into a sequence of values and the second one converts a dictionary into a sequence of key/values. One very frequent use of the unpacking operators is to define a function that can accept any positional and named parameters. That is what I did, because I don't know in advance what the first function in the chain will be, so I have to be prepared to accept any parameters.

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