递归 function 到嵌套列表

Question

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.

Answer 1

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

Answer 2

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.