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pythonrecursionnested-lists
Dr.jj
Dr.jj

Reputation: 31

Recursive function to nested lists

I have tried to crate a recursive function that gets a number and returns a list of nested list according to the index. For example, for 0, the function returns an empty list []. for 1, the function returns [[], [[]]]. for 3 the function returns [ [] , [ [] ] , [ [] , [ [] ] ]], and so on.

def func(n):
    if n == 0:
        return []
    return [func(n-1)]

i have tried many different approaches for this problem. I cant figure out how to extend my output to nested list according to the task.

Upvotes: 2

Views: 256

Answers (2)

chepner
chepner

Reputation: 531125

Each list actually contains all the preceding lists, not just the immediately preceding list. (Based on your example for func(3), your question seems to mistakenly refer to the list returned by func(2) as the list returned by func(1).)

func(0) == []
func(1) == [func(0)] == [[]]
func(2) == [func(0), func(1)] == [[], [[]]]
func(3) == [func(0), func(1), func(2)] == [[] , [[]] , [[], [[]]]]
...
func(n) == [func(0), func(1), ..., func(n-1)]

This is basically a set-theoretic definition of the natural numbers, due to von Neumann. Zero is define to be the empty set, and the successor of each number x is the union of x and the set containing x:

0 == {}
1 == 0 | {0} == {} | {{}} == {{}}
2 == 1 | {1} == {{}} | {{{}}} == {{}, {{}}}

I leave it as an exercise to implement this using lists instead of sets.

Upvotes: 1

Timmy Diehl
Timmy Diehl

Reputation: 309

What I think you're looking for is something like this:

def f(n):
    L = []
    for i in range(n):
        L.append(f(i))
    return L

My confusion stems from your examples. For n=0, there are 0 elements and for n=3 there are 3 elements, but there are 2 elements for n=1. This should work where n is the number of elements in the final list

Upvotes: 1

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