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'
                           }
                   } 

Answer 1

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'

Answer 2

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.