Python list recursion

I'm also not sure what you want to happen, but if you want to split a string into its characters and put them into nested lists, this function works.

def recursion(l,s):
    if s == "":
        return l;
    elif len(l) == 0:
        nL = [s[0]]
        return recursion(nL,s[1:])
    else:
        nL = [l,s[0]]
        return recursion(nL,s[1:])

So for example

print recursion([],"spam")

would output

[[[['s'], 'p'], 'a'], 'm']

I'm not sure what you want to happen, but here's what is happening:

def recursion(x,i):
   x.append(list('spam'))
   x=[x]

Here, x becomes [['s','p','a','m']]. But it doesn't change after this. When you call recursion(x,i) a few lines later, this does not affect the original value of x.

Maybe if you do x = recursion(x,i) instead it will give you what you want, because x will actually be change at the top level.

I think you might be confused by the fact that a Python string ("spam" is a string) is in many ways equivalent to a Python list. You can index them, get their len(), etc. In particular, pretty much anything you do in square brackets works for both string and list types.

Your first example is of a single-byte string, [[['a'], 'a'], 'a'] but you don't give us some key details: what was the input that you expected to produce this output?

For example, the input might have been:

func('a', 3)

Or might have been:

func('aaa')

Or might have been:

func(['a'], 3)   # or ,2)

Or might have been:

func(['a', 'a', 'a'])

Any of those would be a reasonable starting point for a return value of [[['a'], 'a'], 'a'].

Solution

Because your example function takes a second parameter, i, I'm going to assume there is a real need for it. So let's go with the simplest possible case:

def func(lst, times):
    """
    Given a list and a multiplier > 0, return a "nested" list with 
    the contents of the original list repeated (nested) that many
    times:

        Input:  [a,b],2
        Output: [[a,b],a,b]
    """
    assert times > 0
    if times == 1:
        return lst[:]
    else:
        result = lst[:]
        result.insert(0, func(lst, times-1))
        return result

    for times in range(1,4):
        print(func(['a', 'b'], times))

Alternate

Here's a simpler function, that doesn't assume a list:

def func(content, times):
    """
    Given content and a multiplier > 0, return a set of nested lists,
    with the requested depth, each containing optionally any further
    nested lists and the given content.

        Input:  'content', 2
        Output: [['content'], 'content']
    """
    assert(times > 0)
    if times == 1:
        result = [content]
    else:
        result = [func(content, times-1)]
        result.append(content)
    return result

for times in range(1,4):
    print(func('a', times))