Recursive function on sublist returning None

Question

I\'m running a recursive function on sublist to search the element check_value in the list once it finds it, it verifies whether other_value is the first item of the correspondi

Answer 1

I made this code that is quite similar to yours: Instead od using lists, I used dictionary to recursively mark where you can find value, then I made same with list + tuples.

import pprint


def check_with_list(dd, check_value):
    my_indexes = {}
    for index, h in enumerate(dd):
        if isinstance(h, list):
            result = check_with_list(h, check_value)

            if result is not None:
                my_indexes[index] = result
        elif h == check_value:
            my_indexes[index] = True
    return my_indexes


def check_with_list_2(dd, check_value):
    my_indexes = []
    for index, h in enumerate(dd):
        if isinstance(h, list):
            result = check_with_list_2(h, check_value)

            if result is not None:
                my_indexes.append((index, result))
        elif h == check_value:
            my_indexes.append(index)
    return my_indexes


dd = [
    "aaa",
    "bbb",
    ["bbb", "ccc", "bbb"],
    ["bbb", ["ccc", "aaa", "bbb"], "aaa"]
]

rv = check_with_list(dd, "bbb")  # (1,2(0,2),3(0,1(2)))
pprint.pprint(rv)
rv = check_with_list_2(dd, "bbb")  # (1,2(0,2),3(0,1(2)))
pprint.pprint(rv)

Returned values

{1: True, 2: {0: True, 2: True}, 3: {0: True, 1: {2: True}}}
[1, (2, [0, 2]), (3, [0, (1, [2])])]