Searching a nested list with recursion

Question

I want to check if a certain number exists in a nested list

Slist = [[[10, 20], [30, 40]], [[50, 60], [70, 80]]]

def ASearch(t, L):

if t==L:
    return True
else:

    return  ASearch(t, L[0]) or ASearch(t, L[1])

ASearch(15,Slist)

What is the problem?

Update:

The OP meant to post this:

Slist = [[[10, 20], [30, 40]], [[50, 60], [70, 80]]]

def ASearch(t, L):

    if t==L:
        return True
    return ASearch(t, L[0]) or ASearch(t, L[1])


print(ASearch(15, Slist))
print(ASearch(50, Slist))

Expected output:

False
True

Actual output:

TypeError: 'int' object is not subscriptable

15 is not in the list so it will never return True? Your indentation is wrong? — Alan Kavanagh, Nov 17 '17 at 10:54
Can 2 'or'd values ever evaluate to 15? Will it always turn x or y as a True/False? — Alan Kavanagh, Nov 17 '17 at 10:56
I think the [second answer here](https://stackoverflow.com/questions/40514139/check-if-an-item-is-in-a-nested-list) will do what you want. Using the any keyword also has the advantage of working with lists of any length, rather than explicitly slicing the list like you do above. — ConorSheehan1, Nov 17 '17 at 11:00
@user8956966: Please make sure you post with correct indentation. If that is the identical indentation you used in your code then please also fix it in your code! — mrCarnivore, Nov 17 '17 at 11:06

Julien · Accepted Answer · 2017-11-17T11:04:06.683

2

One of the few problems in your code is that it can only return True.

What you want is probably this:

def ASearch(t, L):
    if type(L) == list:
        return any(ASearch(t, l) for l in L)
    else:
        return t == L

edited Nov 17 '17 at 11:04

answered Nov 17 '17 at 11:00

Julien

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[Comment deleted because it was wrong and misleading.] – mrCarnivore Nov 17 '17 at 11:01
@mrCarnivore example? – Julien Nov 17 '17 at 11:01
I must take back my comment after trying it out myself. It does indeed work. Apparently I overlooked the nested call to ASearch! – mrCarnivore Nov 17 '17 at 11:03
@mrCarnivore You can always delete your comment completely. Next to your name and comment timestamp, there is a subtle delete button :-) – Mr. Xcoder Nov 17 '17 at 11:20
@Mr.Xcoder: I know. But then the other users comment does not make any sense any more. And I did not want to cause an avalanche. :-) – mrCarnivore Nov 17 '17 at 11:28

score 0 · Answer 2 · answered Nov 17 '17 at 11:13

Another approach would be flattening the list separately and then checking if the element is included in the flattened list:

def flatten(L):
    flat = []
    if isinstance(L, list):
        for item in L:
            flat += flatten(item)
    else:
        flat.append(L)
    return flat


def ASearch(t, L):
    return t in flatten(L)

You can now call your ASearch function using the element and the list: ASearch(50, nested_list).

Alternatively, you can make ASearch a lambda-function now:

ASearch = lambda t,L: t in flatten(L)

Searching a nested list with recursion

2 Answers2