How to compose a list of functions in steps in Python using map and reduce

Question

Given a list of functions (functions) and an integer n, I'm trying to figure out a way to compose them stepwise, and return a list of each stepwise result as follows:

compose_step([lambda x: x+3, lambda x: x+5, lambda x: x+1], 8) --> [8, 11, 16, 17]

So, as of now I have figured out how to compose a list of functions and return the result as follows:

def compose(functions, n):
    def compose2(f,g):
        return lambda x: f(g(x))
    composedFunction = functools.reduce(compose2, functions, lambda x: x)
    return composedFunction(n)

However, I'm extremely confused how to keep track of each step and return it as a list. I'm assuming that I need to use map somehow in order to map each stepwise piece to the list. I have also come up with a method to apply all functions in the list to n as such:

def apply_all_functions(functions, n):
    answerList = list(map(methodcaller('__call__', n), functions)))
    return answerList

I was thinking of somehow using the composeFunction function to compose a new list of stepwise functions, all the way up until the fully composed function, and then using this as my new list for the apply_all_functions in order to achieve the desired result. But currently, I'm pretty stumped.

Can you explain more what is your aim here? I don't understand what is your input and expected output. — anishtain4
– anishtain4, Commented Apr 24, 2018 at 22:30

Patrick Haugh · Accepted Answer · 2018-04-24 22:32:17Z

4

You can use itertools.accumulate with a composition function

from itertools import accumulate

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

funcs = [lambda x: x, lambda x: x+3, lambda x: x+5, lambda x: x+1]

print([f(8) for f in accumulate(funcs, compose)])
# [8, 11, 16, 17]

answered Apr 24, 2018 at 22:32

Patrick Haugh

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Comments

Mulan · Accepted Answer · 2018-04-25 16:33:05Z

itertools.accumulate is the way to go, but if you're wondering how you could do it on your own, here's one way

def apply_all (x, f = None, *fs):
  if f is None:
    return []
  else:
    next = f (x)
    return [ next ] + apply_all(next, *fs)

funcs = \
  [ lambda x: x
  , lambda x: x+3
  , lambda x: x+5
  , lambda x: x+1
  ]

print(apply_all(8, *funcs))
# [ 8, 11, 16, 17 ]

If you need the form in your original question

def apply_all (fs, x):
  if not fs:
    return []
  else:
    next = fs[0](x)
    return [ next ] + apply_all(fs[1:], next)

funcs = \
  [ lambda x: x
  , lambda x: x+3
  , lambda x: x+5
  , lambda x: x+1
  ]

print(apply_all(funcs, 8))
# [ 8, 11, 16, 17 ]

The above form operates on fs[0] and fs[1:] which shows this can be expressed as a vanilla reduce

from functools import reduce

def apply_all (fs, x):
  def reducer (acc, f):
    (seq, x) = acc
    next = f (x)
    return (seq + [next], next)
  return reduce(reducer, fs, ([], x)) [0]

funcs = \
  [ lambda x: x
  , lambda x: x+3
  , lambda x: x+5
  , lambda x: x+1
  ]

print(apply_all(funcs, 8))
# [ 8, 11, 16, 17 ]

Ogaday · Accepted Answer · 2022-11-14 15:58:41Z

0

You can also use itertools.accumulate with an initial value:

from itertools import accumulate


def compose(data, funcs):
    yield from accumulate(funcs, lambda data, f: f(data), initial=data)


funcs = [lambda x: x+3, lambda x: x+5, lambda x: x+1]
init = 8
print(list(compose(init, funcs)))
# [8, 11, 16, 17]

You can do something similar with functools.reduce if you don't want the intermediate results.

This is similar to @Patrick's answer, but here accumulate applies each function and yields the stepwise result one by one instead of generating composed functions.

answered Nov 14, 2022 at 15:58

Ogaday

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