Returning keys nested dictionary with Python

Question

EDIT: Thanks to @arzo for the catch

I have a nested dictionary structured as:

{
"key1":"string",
"key2":[{"nestedkey":"nestedvalue"}],
"key3":[1,2,3],
"key4":[{"nestedlevel1key":[{"nestedlevel2key":"nestedlevel2value"}]}],
"key5": {},
"key6": {"regularkey": "regularvalue"},
"key7": 15
}

In which a dictionary key can contain:

1. strings
2. ints
3. lists of dictionaries
4. lists of dictionaries that contain lists of dictionaries
5. an empty dictionary
6. a regular dictionary

The problem statement is optimizing a method that can handle returning all of the keys within a dictionary. I can write something like this:

def get_keys(dict_example):
    keys = []

    for k,v in dict_example.items():
        keys.append(k)
        if isinstance(v, dict):
            for k in v.keys():
                keys.append(k)
        if isinstance(v, list):
            if isinstance(v[0], dict):
                for k,v in v[0].items():
                    keys.append(k)
                    if isinstance(v, list) and isinstance(v[0], dict):
                        for k in v[0].keys():
                            keys.append(k) 
    return keys
         keys = get_keys(dict_example)

            
print(keys)

Which will get me (in no particular order) a list of the keys:

['key1', 'key2', 'nestedkey1', 'key3', 'key4', 'nestedlevel1key', 'nestedlevel2key', 'key5', 'key6', 'regularkey', 'key7']

But I am not sure of an optimized method that is simplified to take on all the 6 use cases that can also transverse through the array, regardless of how many levels there are. Now I made the heuristic of the number of levels, but there could be deeper levels within this array that I need to account for.

Answer 1

They are only 2 cases to handle: list and dict as they contain other thing, then use recursivity

• for a dict : keeps keys, and search throught values
• for list : search throught values

def get_keys(item):
    keys = []
    if isinstance(item, dict):
        for k, v in item.items():
            keys.append(k)
            keys.extend(get_keys(v))
    elif isinstance(item, (list, tuple)):
        for x in item:
            keys.extend(get_keys(x))
    return keys

---

values = {
    "key1": "string", "key2": [{"nestedkey": "nestedvalue"}], "key3": [1, 2, 3],
    "key4": [{"nestedlevel1key": [{"nestedlevel2key": "nestedlevel2value"}]}],
    "key5": {}, "key6": {"regularkey": "regularvalue"},
}

keys = get_keys(values)
print(keys)
# ['key1', 'key2', 'nestedkey', 'key3', 'key4', 'nestedlevel1key', 'nestedlevel2key', 'key5', 'key6', 'regularkey']

Answer 2

Rather than a high number of nested loops, you can solve this with a recursive function:

origDict = {
"key1":"string",
"key2":[{"nestedkey":"nestedvalue"}],
"key3":[1,2,3],
"key4":[{"nestedlevel1key":[{"nestedlevel2key":"nestedlevel2value"}]}],
"key5": {},
"key6": {"regularkey": "regularvalue"},
"key7": 15
}

keyList = []

def get_keys(inDict, listOfKeys):
    for k, v in inDict.items():
        listOfKeys.append(k)
        if isinstance(v, list):
            for item in v:
                if isinstance(item, dict):
                    get_keys(item, listOfKeys)
        elif isinstance(v, dict):
            get_keys(v, listOfKeys)

get_keys(origDict, keyList)
print(keyList)

This prints:

['key1', 'key2', 'nestedkey', 'key3', 'key4', 'nestedlevel1key', 
'nestedlevel2key', 'key5', 'key6', 'regularkey', 'key7']```

Answer 3

Just for fun, you can also use a non-recursive function to save time:

def get_keys(d):
    out = []
    inside = d
    while inside:
        out.append(list(inside.keys())[0])
        v = inside.pop(out[-1])
        if isinstance(v, dict):
            inside.update(v)
        elif isinstance(v, list) and isinstance(v[0], dict):
            for x in v:
                inside.update(x)
        else:
            continue
    return out

out = get_keys(d)

Output:

['key1', 'key2', 'key3', 'key4', 'key5', 'key6', 'key7', 'nestedkey', 'nestedlevel1key', 'regularkey', 'nestedlevel2key']