remove multiple layers of lists in a list without using library

Question

I want to make a function that removes any layers of lists in a list

Example

[[[[(1,2)]]],[[(3,4)]]] --> [(1,2),(3,4)]

[(1, 2), [(1, 2)]] -->[(1,2),(1,2)]

Here is my attempt using recursion which didnt work on the second example.

def nolistlist(t): 
    flat_list = []
    for i in t:
        if isinstance(i,list):
            for x in t:   
                if type(x) == list:
                    for y in x:
                        flat_list.append(y)
                else:
                    flat_list.append(x)
            return nolistlist(flat_list)
        else:
            return t

Anyone that has a simple solution without importing any library? I know the problem but cant come up with a solution. Have been trying to solve this problem for while now. Help would be appreciated!

You could use recursion if you don't know how many layers the list you're given will be. — nonamorando, Mar 12 '21 at 01:10
`numpy.flatten` can do it, but of course that violates your "no library" rule. — Tim Roberts, Mar 12 '21 at 01:14
A oneliner: `[*eval(str(t).replace('[', '').replace(']', ''))]` — superb rain, Mar 12 '21 at 01:31

score 3 · Answer 1 · answered Mar 12 '21 at 01:22

3

Another version, without using global variable:

l = [(1, 2), [(1, 2)]]


def flatten(l):
    if isinstance(l, list):
        for v in l:
            yield from flatten(v)
    else:
        yield l

print(list(flatten(l)))

Prints:

[(1, 2), (1, 2)]

answered Mar 12 '21 at 01:22

Andrej Kesely

168,389
15
48
91

Better. I just didn't see it. – Tim Roberts Mar 12 '21 at 01:23
@TimRoberts Yours is more efficient for deeply nested lists, though. – superb rain Mar 12 '21 at 01:32
Dumb question, never seen ``yield`` before. How do i return the result? Tried ``return list(flatten(v))`` but it didnt work.. – Che Che Mar 12 '21 at 01:32
`yield` returns a result, AND lets the function continue on to return more results. It's a generator. You don't need a return. The code works as is. – Tim Roberts Mar 12 '21 at 05:19

score 1 · Answer 2 · answered Mar 12 '21 at 01:19

1

I hate this answer because it relies on a global, but I haven't figured out how to solve it with nested yields yet.

x = [[[[(1,2)]]],[[(3,4)]]] #--> [(1,2),(3,4)]

glist = []
def notlist(z):
    for i in z:
        if isinstance(i,list):
            notlist(i)
        else:
            glist.append(i)
    return glist

print( notlist(x) )

answered Mar 12 '21 at 01:19

Tim Roberts

48,973
4
21
30

for some reason my program takes alot more time when i use global... – Che Che Mar 12 '21 at 01:36

remove multiple layers of lists in a list without using library

2 Answers2