Sequence of nested lists recursive python

Question

Given some number n, I want to produce a list of size n where the following example shows how the n'th element in the list should be:

For n=0: return []
For n=1: return [[]]
For n=2: return [[],[[]]]
For n=3: return [[], [[]], [[], [[]]]

Basically it takes the precedent lists and append them into a new list.

I've tried the following:

def magic_list_helper(n, lst):
    if n == 0:
        return lst
    if n == 1:
        lst.append([])
        return magic_list_helper(n-1,lst)
    if n > 1:
        magic_list_helper(n-1, lst)
        new_list = []
        new_list.append(lst[n - 2])
        lst.append(new_list)
        return lst

but for n=3 I get [[], [[]], [[[]]]].

Does this answer your question? Recursive function to nested lists — Timmy Diehl
– Timmy Diehl, Commented Dec 12, 2022 at 21:19
See also Generate increasing nested empty list structure with recursion which is a closer spec match than Recursive function to nested lists but doesn't have an upvoted answer yet. — ggorlen
– ggorlen, Commented Dec 12, 2022 at 22:12

Lexpj · Accepted Answer · 2022-12-12 21:25:11Z

1

There are a few things to notice. First, how will you set your recursion up in a way that produces said output. Well, your output is of the form [f(0), f(1), f(2) ... f(n)] for recursive function f. The following does exactly that:

def magic_list_helper(n):
    return [] if n == 0 else [magic_list_helper(n-i) for i in range(1,n)]

If n=0, it returns an empty list, else, it is going to return [f(0), f(1), f(2) ... f(n)], and this, will then return [f(0), f(1), f(2) ... f(n-1)] and so on.

answered Dec 12, 2022 at 21:25

Lexpj

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2 Comments

Le S Over a year ago

Im sorry I forgot to add that I cant use for loops

ggorlen Over a year ago

Any for loop can be refactored to recursion or a while loop (not that it's a good idea in general, just to fulfill arbitrary requirements).

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