Python - Unflatten dict

Question

I have this multi-dimensional dict:

a = {'a' : 'b', 'c' : {'d' : 'e'}}

And written simple function to flatten that dict:

def __flatten(self, dictionary, level = []):
    tmp_dict = {}
    for key, val in dictionary.items():
        if type(val) == dict:
            tmp_dict.update(self.__flatten(val, level + [key]))
        else:
            tmp_dict['.'.join(level + [key])] = val
    return tmp_dict

After call this function with dict a i get in result:

{'a' : 'b', 'c.d' : 'e'}

Now, after making few instructions on this flattened dict i need to build new, multi-dimensional dict from that flattened. Example:

>> unflatten({'a' : 0, 'c.d' : 1))
{'a' : 0, 'c' : {'d' : 1}}

The only problem I have is that i do not have a function unflatten :)
Can anyone help with this? I have no idea how to do it.

EDIT:

Another example:

{'a' : 'b', 'c.d.e.f.g.h.i.j.k.l.m.n.o.p.r.s.t.u.w' : 'z'}

Should be after unflatten:

{'a': 'b', 'c': {'d': {'e': {'f': {'g': {'h': {'i': {'j': {'k': {'l': {'m': {'n': {'o': {'p': {'r': {'s': {'t': {'u': {'w': 'z'}}}}}}}}}}}}}}}}}}}

---

And another:

{'a' : 'b', 'c.d' : 'z', 'c.e' : 1}

To:

{'a' : 'b', 'c' : {'d' : 'z', 'e' : 1}}

This greatly increases the difficulty of the task, i know. Thats why i had problem with this and found no solution in hours..

Answer 1

def unflatten(dictionary):
    resultDict = dict()
    for key, value in dictionary.items():
        parts = key.split(".")
        d = resultDict
        for part in parts[:-1]:
            if part not in d:
                d[part] = dict()
            d = d[part]
        d[parts[-1]] = value
    return resultDict

Answer 2

from collections import defaultdict
def unflatten(d):
    ret = defaultdict(dict)
    for k,v in d.items():
        k1,delim,k2 = k.partition('.')
        if delim:
            ret[k1].update({k2:v})
        else:
            ret[k1] = v
    return ret

Answer 3

Here's one utilizing Python 3.5+ features, like typing and destructuring assignments. Try the tests out on repl.it.

from typing import Any, Dict


def unflatten(
    d: Dict[str, Any], 
    base: Dict[str, Any] = None,
) -> Dict[str, Any]:
    """Convert any keys containing dotted paths to nested dicts

    >>> unflatten({'a': 12, 'b': 13, 'c': 14})  # no expansion
    {'a': 12, 'b': 13, 'c': 14}

    >>> unflatten({'a.b.c': 12})  # dotted path expansion
    {'a': {'b': {'c': 12}}}

    >>> unflatten({'a.b.c': 12, 'a': {'b.d': 13}})  # merging
    {'a': {'b': {'c': 12, 'd': 13}}}

    >>> unflatten({'a.b': 12, 'a': {'b': 13}})  # insertion-order overwrites
    {'a': {'b': 13}}

    >>> unflatten({'a': {}})  # insertion-order overwrites
    {'a': {}}
    """
    if base is None:
        base = {}

    for key, value in d.items():
        root = base

        ###
        # If a dotted path is encountered, create nested dicts for all but
        # the last level, then change root to that last level, and key to
        # the final key in the path.
        #
        # This allows one final setitem at the bottom of the loop.
        #
        if '.' in key:
            *parts, key = key.split('.')

            for part in parts:
                root.setdefault(part, {})
                root = root[part]

        if isinstance(value, dict):
            value = unflatten(value, root.get(key, {}))

        root[key] = value

    return base

Answer 4

I wrote one years ago in Python 2 and 3 which I've adapted below. It was for making it easier to check if a given dictionary is a subset of a larger dictionary irrespective of whether provided in flattened or scaffolded form.

A bonus feature: Should there be consecutive integer indexes (as in 0, 1, 2, 3, 4 etc.), this will also convert them back into lists as well.

def unflatten_dictionary(field_dict):
    field_dict = dict(field_dict)
    new_field_dict = dict()
    field_keys = list(field_dict)
    field_keys.sort()

    for each_key in field_keys:
        field_value = field_dict[each_key]
        processed_key = str(each_key)
        current_key = None
        current_subkey = None
        for i in range(len(processed_key)):
            if processed_key[i] == "[":
                current_key = processed_key[:i]
                start_subscript_index = i + 1
                end_subscript_index = processed_key.index("]")
                current_subkey = int(processed_key[start_subscript_index : end_subscript_index])

                # reserve the remainder descendant keys to be processed later in a recursive call
                if len(processed_key[end_subscript_index:]) > 1:
                    current_subkey = "{}.{}".format(current_subkey, processed_key[end_subscript_index + 2:])
                break
            # next child key is a dictionary
            elif processed_key[i] == ".":
                split_work = processed_key.split(".", 1)
                if len(split_work) > 1:
                    current_key, current_subkey = split_work
                else:
                    current_key = split_work[0]
                break

        if current_subkey is not None:
            if current_key.isdigit():
                current_key = int(current_key)
            if current_key not in new_field_dict:
                new_field_dict[current_key] = dict()
            new_field_dict[current_key][current_subkey] = field_value
        else:
            new_field_dict[each_key] = field_value

    # Recursively unflatten each dictionary on each depth before returning back to the caller.
    all_digits = True
    highest_digit = -1
    for each_key, each_item in new_field_dict.items():
        if isinstance(each_item, dict):
            new_field_dict[each_key] = unflatten_dictionary(each_item)

        # validate the keys can safely converted to a sequential list.
        all_digits &= str(each_key).isdigit()
        if all_digits:
            next_digit = int(each_key)
            if next_digit > highest_digit:
                highest_digit = next_digit

    # If all digits and can be sequential order, convert to list.
    if all_digits and highest_digit == (len(new_field_dict) - 1):
        digit_keys = list(new_field_dict)
        digit_keys.sort()
        new_list = []

        for k in digit_keys:
            i = int(k)
            if len(new_list) <= i:
                # Pre-populate missing list elements if the array index keys are out of order
                # and the current element is ahead of the current length boundary.
                while len(new_list) <= i:
                    new_list.append(None)
            new_list[i] = new_field_dict[k]
        new_field_dict = new_list
    return new_field_dict

# Test
if __name__ == '__main__':
    input_dict = {'a[0]': 1,
                  'a[1]': 10,
                  'a[2]': 5,
                  'b': 10,
                  'c.test.0': "hi",
                  'c.test.1': "bye",
                  "c.head.shoulders": "richard",
                  "c.head.knees": 'toes',
                  "z.trick.or[0]": "treat",
                  "z.trick.or[1]": "halloween",
                  "z.trick.and.then[0]": "he",
                  "z.trick.and.then[1]": "it",
                  "some[0].nested.field[0]": 42,
                  "some[0].nested.field[1]": 43,
                  "some[2].nested.field[0]": 44,
                  "mixed": {
                      "statement": "test",
                      "break[0]": True,
                      "break[1]": False,
                  }}
    expected_dict = {'a': [1, 10, 5],
                     'b': 10,
                     'c': {
                         'test': ['hi', 'bye'],
                         'head': {
                             'shoulders': 'richard',
                             'knees' : 'toes'
                         }
                     },
                     'z': {
                         'trick': {
                             'or': ["treat", "halloween"],
                             'and': {
                                 'then': ["he", "it"]
                             }
                         }
                     },
                     'some': {
                         0: {
                             'nested': {
                                 'field': [42, 43]
                             }
                         },
                         2: {
                             'nested': {
                                 'field': [44]
                             }
                         }
                     },
                     "mixed": {
                         "statement": "test",
                         "break": [True, False]
                     }}
    # test
    print("Input:")
    print(input_dict)
    print("====================================")
    print("Output:")
    actual_dict = unflatten_dictionary(input_dict)
    print(actual_dict)
    print("====================================")
    print(f"Test passed? {expected_dict==actual_dict}")
    

Answer 5

As a rough-draft (could use a little improvement in variable name choice, and perhaps robustness, but it works for the example given):

def unflatten(d):
    result = {}
    for k,v in d.iteritems():
        if '.' in k:
            k1, k2 = k.split('.', 1)
            v = {k2: v}
            k = k1
        result[k] = v
    return result