Get parents keys from nested dictionary

14,848

Solution 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.

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

print(
    breadcrumb(example_dict_with_list, 'value4aa')
)

Which returns

['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'
Share:
14,848
Jym Morrison
Author by

Jym Morrison

Updated on July 23, 2022

Comments

  • Jym Morrison
    Jym Morrison almost 2 years

    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'
                               }
                       } 
    
  • zelusp
    zelusp almost 8 years
    by def find_key(d,key): you mean def find_key(d,value):, correct?
  • mgilson
    mgilson almost 8 years
    @zelusp -- Yes, the name of that variable was ... Unfortunate. :-). Thanks for the suggestion.
  • Blind0ne
    Blind0ne about 3 years
    How to find multiple occurences? For example a list of lists with the key paths. E.g.: [['key4', 'key4a', 'key4ac'], ['key5, 'key5a', 'key5ac']]