Get parents keys from nested dictionary

Question

From the following nested dictionary, how can I get every parent dictionary key of 'value4ac'? By starting the 'value4ac' value, I want to get 'key4', 'key4a', 'Key4ac'.

example_dict = { 'key1' : 'value1',
                 'key2' : 'value2',
                 'key3' : { 'key3a': 'value3a' },
                 'key4' : { 'key4a': { 
                                         'key4aa': 'value4aa',
                                         'key4ab': 'value4ab',
                                         'key4ac': 'value4ac'
                                     },
                            'key4b': 'value4b'
                           }
                   }

Asked By: Jym Morrison

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Source

Answer 1

recursion to the rescue!

example_dict = { 'key1' : 'value1',
                 'key2' : 'value2',
                 'key3' : { 'key3a': 'value3a' },
                 'key4' : { 'key4a': { 'key4aa': 'value4aa',
                                       'key4ab': 'value4ab',
                                       'key4ac': 'value4ac'},
                            'key4b': 'value4b'}
                }

def find_key(d, value):
    for k,v in d.items():
        if isinstance(v, dict):
            p = find_key(v, value)
            if p:
                return [k] + p
        elif v == value:
            return [k]

print find_key(example_dict,'value4ac')

how it works

It looks through the items and checks 2 cases

item isn’t a dictionary — In this case, see if it is the value we’re looking for. If it is, return a list which contains only the key. (This is our base-case for recursion).
item is a dictionary — Try looking for the key in that dictionary. If it is found in that dictionary (or any sub-dict), return the key which takes the right path pre-pended onto the rest of the path.

Answered By: mgilson

Answer 2

What if your dictionary contains lists too?

Here’s a more broad variant of @mgilson ‘s solution, suitable for JSON:

example_dict_with_list = { 'key1' : 'value1',
                           'key2' : 'value2',
                           'key3' : { 'key3a': 'value3a' },
                           'key4' : { 'key4a': [{ 'key4aa': 'value4aa',
                                                  'key4ab': 'value4ab',
                                                  'key4ac': 'value4ac'}],
                                      'key4b': 'value4b'}
                          }

def breadcrumb(json_dict_or_list, value) -> list:
    if value == json_dict_or_list:
        return [json_dict_or_list]
    elif isinstance(json_dict_or_list, str) and isinstance(value, str):
        if value.casefold() in json_dict_or_list.casefold():
            return [json_dict_or_list]
    elif isinstance(json_dict_or_list, dict):
        for k, v in json_dict_or_list.items():
            match = breadcrumb(k, value)
            if match:
                return match + [v]
            match = breadcrumb(v, value)
            if match:
                return [k] + match
    elif isinstance(json_dict_or_list, list):
        lst = json_dict_or_list
        for i in range(len(lst)):
            match = breadcrumb(lst[i], value)
            if match:
                return [str(i)] + match

Test it:

# search by exact value:
print(breadcrumb(example_dict_with_list, 'value4aa'))
# search by a substring:
print(breadcrumb(example_dict_with_list, 'value4a'))
# search by key:
print(breadcrumb(example_dict_with_list, 'key4aa'))

Result:

['key4', 'key4a', '0', 'key4aa', 'value4aa']
['key4', 'key4a', '0', 'key4aa', 'value4aa']
['key4', 'key4a', '0', 'key4aa', 'value4aa']

Bonus

If you need to print it out nicely, like a string of breadcrums, do

print(
  ' > '.join(
    breadcrumb(example_dict, 'value4aa')
  )
)

Which will return

'key4 > key4a > 0 > key4aa > value4aa'

Answered By: KiriSakow

Answer 3

You can use a NestedDict (I am the author).

>>> from ndicts.ndicts import NestedDict
>>> example_dict = { 'key1' : 'value1',
...                  'key2' : 'value2',
...                  'key3' : { 'key3a': 'value3a' },
...                  'key4' : { 'key4a': {
...                                          'key4aa': 'value4aa',
...                                          'key4ab': 'value4ab',
...                                          'key4ac': 'value4ac'
...                                      },
...                             'key4b': 'value4b'
...                            }
...                    }
>>> nd = NestedDict(example_dict)
>>> for key, value in nd.items():
...     if value == "value4ac":
...             print(key)
...
('key4', 'key4a', 'key4ac')

To install ndicts

pip install ndicts

Answered By: edd313

Answer 4

Solution that supports dicts keys and lists elements

Slight modification of @KiriSakow solution (whom was inspired by @mgilson).

It will not produce a list index and returns correct parents not only for the last child, but for sub-parents as well.

def breadcrumb(nested_dict, value):
    if nested_dict == value:
        return [nested_dict]
    elif isinstance(nested_dict, dict):
        for k, v in nested_dict.items():
            if k == value:
                return [k]
            p = breadcrumb(v, value)
            if p:
                return [k] + p
    elif isinstance(nested_dict, list):
        lst = nested_dict
        for i in range(len(lst)):
            p = breadcrumb(lst[i], value)
            if p:
                return p

Input

example_dict_with_list = { 'key1' : 'value1',
                           'key2' : 'value2',
                           'key3' : { 'key3a': 'value3a' },
                           'key4' : { 'key4a': [{ 'key4aa': 'value4aa',
                                                  'key4ab': 'value4ab',
                                                  'key4ac': 'value4ac'}],
                                      'key4b': 'value4b'}
                          }


print(breadcrumb(example_dict_with_list, "value4aa"))
print(breadcrumb(example_dict_with_list, "key4aa"))

Output

['key4', 'key4a', 'key4aa', 'value4aa']
['key4', 'key4a', 'key4aa']

Answered By: Alex Kopylov