Python function chaining an unknown number of times

Question

Is there a way in Python to chain the same function an unknown number of times on an object?

For example. If I wanted to chain the function fn on the object obj twice I'd have:

obj.fn(**kwargs1).fn(**kwargs2)

If I wanted to chain a fn five times I'd have:

obj.fn(**kwargs1).fn(**kwargs2).fn(**kwargs3).fn(**kwargs4).fn(**kwargs5)

Answer 1

You can actually use a simple for loop, which iterates over the list of arguments, and applies each argument to the object, and reassign the results back to the object

for args in list_of_args:

    obj = obj.fn(*args)

For example, I can use this logic to chain string.replace as follows

obj = 'aaaccc'
list_of_args = [['a','b'], ['c','d']]

for args in list_of_args:
    obj = obj.replace(*args)

print(obj)

And the output will be

bbbddd

Which is the same as doing 'aaabbb'.replace('a','b').replace('c','d')

Or we can also perhaps use a recursive method, which takes in the object, and a counter which serves as an index of the current argument to be applied, and a number of times the function needs to run

#fn is either defined, or imported from outside
def chain_func(obj, counter, max_count):

    #If counter reaches max no of iterations, return
    if counter == max_count:
        return obj

    #Else recursively apply the function
    else:
        return chain_func(obj.fn(list_of_args[counter]), counter+1, max_count)

#To call it twice
list_of_args = [kwargs1, kwargs2]
chain_func(obj, 0, 2)

For example, I can use this logic to chain string.replace as follows

def chain_func(obj, counter, max_count):

    #If counter reaches max no of iterations, return
    if counter == max_count:
        return obj

    #Else recursively apply the function
    else:
        return chain_func(obj.replace(*list_of_args[counter]), counter+1, max_count)


list_of_args = [['a','b'], ['c','d']]
print(chain_func('aaaccc', 0, 2))

And the output will be

bbbddd