Recursive function on sublist returning None

0
votes

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 corresponding list and finally return the index.But current code is returning None.can anyone please support as I'm not having much understanding on recursive functions functioning on sublists.

 def check_with_list(dd, check_value, other_value=None):
    global new_index

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

            if result is not None:
                if other_value:
                    new = (index,) + result
                    if len(new) == 2:

                        if not dd[new[0]][0] == other_value:
                            result = None
                        else:
                            return (index,) + result


        elif h == check_value:
            return (index,)
    # value not found
    return None


dd = [
    "gcc",
    "fcc",
    ["scc", "jhh", "rrr"],
    ["www", "rrr", "rrr"],
    "mmm",
    ["qwe", ["ree", "rer", "rrr"], "ere"]
]
dd = check_with_list(dd, "rrr", "ree")

print(dd)
3
pythonrecursion
What do you want the answer to be? - user2653663
2,2 if 'scc' is passed as other_value whereas 3,1 when 'www' is the other_value. - Ashwin kumar
Does this need to be recursive? Can there be further levels of sub-lists? - AKX
Yes, there can be further lists. - Ashwin kumar
if 'ree' is passed as other value index should be 5,1,2. - Ashwin kumar

3 Answers

1
votes
def check_with_list(dd, check_value, other_value=None):
        global new_index
    for index, h in enumerate(dd):
        if isinstance(h, list):
            result = check_with_list(h, check_value)

            if result is not None:
                if other_value:
                    new = (index,) + result
                    if len(new) == 2:

                        if dd[new[0]][0] == other_value:
                            result = None
                        else:
                            return (index,) + result


        elif h == check_value:
            return (index,)
    # value not found
    return None


dd = [
    "gcc",
    "fcc",
    ["scc", "jhh", "rrr"],
    ["www", "rrr", "rrr"],
    "mmm",
    ["qwe", ["ree", "rrr", "rrr"], "ere"]
]
dd = check_with_list(dd, "rrr", "ree")

I have removed the not from the line below:

if not dd[new[0]][0] == other_value:

Everything else seems to be perfect. The code works and returns the index of the 1st occurrence of check_value in dd.

0
votes

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])])]
0
votes

I believe you structured your logic incorrectly, looking for other_value after the opportunity has passed. Here's an alternate way to structure this:

def check_with_list(structure, check_value, other_value=None):

    for index, item in enumerate(structure):
        path = (index,)

        if isinstance(item, list):

            sub_path = check_with_list(item, check_value, other_value)

            if sub_path is not None:

                path += sub_path

                if other_value and check_value in item:

                    if item[0] == other_value:
                        return path
                else:
                    return path

        elif item == check_value:
            return path

    return None  # value not found

dd = [
    "gcc",
    "fcc",
    ["scc", "jhh", "rrr"],
    ["www", "rrr", "rrr"],
    "mmm",
    ["qwe", ["ree", "rer", "rrr"], "ere"]
]

print(check_with_list(dd, "rrr", "ree"))

OUTPUT

> python3 test.py
(5, 1, 2)
>