How to chain Python function calls so the behaviour is as follows

Question

I stumbled upon the following problem on a python challenge: Write a function that satisfies the following rule for any number of function calls.

f()()()()()(s) == 'fooooo' + s;

example:

f('it') == 'fit';
f()('x') == 'fox';
f()()('bar') == 'foobar';
f()()()('l') == 'foool';

The function should be stateless and should not use any variables outside the scope.

The function signature was:

def f(s=None):
    # Your code here

I thought that in order to be able to chain multiple calls we will have to return a function when no string is passed into the function, but can't figure out how to build the expected string with no external variables. Suggestions?

def f(s=None):
    if s is None:
        # concatenate an 'o'?
        return f
    else:
        # Last call, return the result str.
        return s

sloth · Accepted Answer · 2019-06-03 11:47:01Z

22

An alternative to Nikola's answer is something like this:

def f(s=None):
    if s: return f'f{s}'

    def factory(prefix):
        def inner(s=None):
            return f'f{prefix}{s}' if s else factory(prefix + 'o')
        return inner
    return factory('o')

using a closure and no helper function.

edited Jun 3, 2019 at 11:47

answered Jun 3, 2019 at 11:33

sloth

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Comments

Nikola Dimitroff · Accepted Answer · 2019-06-03 11:26:09Z

13

Obviously, you need to store the number of 'o' somewhere in the memory (e.g. the code) of f. To achieve this, you can benefit from these 2 features of Python:

You can define functions inside other functions
There's this thing called argument binding which allows you to tell Python to fix the some or all of the arguments of your function. This is done through functools.partial

And here's the solution

from functools import partial

def f(s=None):
    # Define a new function g which takes the current text and takes s
    def g(current_text, s=None):
        if s is not None:
            return current_text + s
        else:
            # If called with an empty argument, just rebind current_text with an extra o
            return partial(g, current_text + "o")

    # Just call g with the initial conditions
    return g("f", s)

print(f()()()()()("s")) # fooooos
print(f("s")) # fs

answered Jun 3, 2019 at 11:26

Nikola Dimitroff

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6 Comments

user8426627 Over a year ago

you are using a class, that is defined outside of a scope

Netwave Over a year ago

@user8426627, no, he is using a helper, but the function data lives whithin the scope.

sloth Over a year ago

@Netwave Well, actually, partial is indeed a class, and you could argue that it's also outside the scope. OTOH the call to partial could be trivially replaced with another nested function and a closure.

Netwave Over a year ago

The function should be stateless and should not use any variables outside the scope., this do not mean that he cannot use any other resource, but that it should live in the function. At least is what I understood.

Javi Over a year ago

Brilliant solution, in my opinion fits the problem requirements!

|

Olvin Roght · Accepted Answer · 2019-06-03 11:52:03Z

4

You can try this:

def f(s=None):
    string = "f"
    def ret(p=None):
        nonlocal string
        string += "o"
        return ret if not p else string + p
    return ret if not s else string + s

edited Jun 3, 2019 at 11:52

answered Jun 3, 2019 at 11:33

Olvin Roght

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3 Comments

sloth Over a year ago

OP said the function signature has to be def f(s=None):, so I don't know if this counts...

Olvin Roght Over a year ago

@sloth, sorry, I've missed that. Updated.

Javi Over a year ago

Brilliant solution, although it could be argued the stateless condition of it.

GBrandt · Accepted Answer · 2019-06-03 11:58:52Z

4

This is my go at it:

def f(x=None, seq=''):
    if x:
        return 'f' + seq + x
    else:
        def g(y=None, p=seq + 'o'):
            return f(y, p)
        return g

Edit If you really need the function signature to be f(x=None), use this:

def f(x=None):
    def f_(x=None, seq=''):
        if x:
            return 'f' + seq + x
        else:
            def g(y=None, p=seq + 'o'):
                return f_(y, p)
            return g
    return f_(x)

:^)

edited Jun 3, 2019 at 11:58

answered Jun 3, 2019 at 11:55

GBrandt

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6 Comments

sloth Over a year ago

OP said the function signature has to be def f(s=None):, so I don't know if this counts...

GBrandt Over a year ago

That's not really a problem though, you can just wrap it into another function

Olvin Roght Over a year ago

or use nonlocal which have been implemented for this :D

GBrandt Over a year ago

But a variable, nonlocal or not, counts as state, doesn't it?

Olvin Roght Over a year ago

Don't know actually, depends on how to count.

|

gehbiszumeis · Accepted Answer · 2019-06-06 07:11:29Z

2

Just for printing the string:

def f(s=None):
    def fo(s=None):
        if s==None:
            print('o',end='')
            return fo
        else:
            print(s)
            return
    if s!=None:
        print('f',end='')
        print(s)
        return
    elif s==None:
        print('fo',end='')
        return fo

edited Jun 6, 2019 at 7:11

gehbiszumeis

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answered Jun 6, 2019 at 6:34

Anay Khator

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Comments

fpbhb · Accepted Answer · 2019-06-09 21:20:26Z

1

Cute problem. This is a compact way to do it:

def f(s=None, prefix="f"):
    if s: return prefix + s
    return lambda s=None: f(s, prefix=prefix+"o")

answered Jun 9, 2019 at 21:20

fpbhb

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Comments

DevineLiu · Accepted Answer · 2019-06-10 03:51:09Z

-1

FP:

f=lambda s=None,prefix='f':prefix+s  if s else lambda s=None,prefix=prefix+'o':f(s,prefix)

answered Jun 10, 2019 at 3:51

DevineLiu

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