CODEWITHSUNDEEP
pythondictionarynested
redcrow
redcrow

Reputation: 1823

Get all values from nested dictionaries in python

I have some dictionaries of dictionaries, like this:

a['b']['c']['d']['answer'] = answer1
a['b']['c']['e']['answer'] = answer2
a['b']['c']['f']['answer'] = answer3
....
a['b']['c']['d']['conf'] = conf1
a['b']['c']['e']['conf'] = conf2
a['b']['c']['f']['conf'] = conf3

Is there a fast way to get a list of values of all answers for all elements at the third level (d,e,f)?

Specifically I'd like to know if there's any mechanism implementing a wildcard (e.g., a['b']['c']['*']['answer'].values()

update The fastest way I've found till now is:

[x['answer'] for x in a['b']['c'].values()]

Upvotes: 22

Views: 39149

Answers (7)

PatrickT
PatrickT

Reputation: 10540

This is too long for a comment. Richard's generator approach to obtain a list of values can also be used to build a list of keys.

def get_nested_values(d):
  for v in d.values():
    if isinstance(v, dict):
      yield from get_nested_values(v)
    else:
      yield v

def get_nested_keys(d):
  for k,v in d.items():
    if isinstance(v, dict):
      yield from get_nested_keys(v)
    else:
      yield k

# get the nested values
list(get_nested_values(d))

# get the nested keys
list(set(get_nested_keys(d)))

Upvotes: 0

Richard
Richard

Reputation: 61539

In Python3 we can build a simple generator for this:

def NestedDictValues(d):
  for v in d.values():
    if isinstance(v, dict):
      yield from NestedDictValues(v)
    else:
      yield v

a={4:1,6:2,7:{8:3,9:4,5:{10:5},2:6,6:{2:7,1:8}}}
print(list(NestedDictValues(a)))

The output is:

[1, 2, 3, 4, 5, 6, 7, 8]

which is all of the values.

Upvotes: 32

edd313
edd313

Reputation: 1477

You can use a NestedDict. First, let me recreate your dictionary

>>> from ndicts.ndicts import NestedDict
>>> nd = NestedDict.from_product("b", "c", "def", ["answer", "conf"])
NestedDict({
    'b': {
        'c': {
            'd': {'answer': None, 'conf': None}, 
            'e': {'answer': None, 'conf': None}, 
            'f': {'answer': None, 'conf': None}
         }
     }
})

Then use an empty string as a wildcard

>>> nd_extract = nd.extract["b", "c", "", "answer"]
>>> nd_extract
NestedDict({
    'b': {
        'c': {
            'd': {'answer': None}, 
            'e': {'answer': None}, 
            'f': {'answer': None}
        }
    }
})

Finally get the values

>>> list(nd_extract.values())
[None, None, None]

To install ndicts

pip install ndicts

Upvotes: 0

Jerry Wong
Jerry Wong

Reputation: 1

list(map(lambda key:  a['b']['c'][key],  a['b']['c'].keys()))

Upvotes: 0

redcrow
redcrow

Reputation: 1823

Just to give an answer to this topic, copying my solution from the "updating status" of my question:

[x['answer'] for x in a['b']['c'].values()]

Hope this can help.

Upvotes: 1

Ankur Ankan
Ankur Ankan

Reputation: 3066

You could use a simple list comprehension:

[a['b']['c'][key]['answer'] for key in a['b']['c'].keys()]
Out[11]: ['answer1', 'answer2', 'answer3']

If you want to get all the answers and conf etc. You could do:

[[a['b']['c'][key][type] for key in a['b']['c'].keys()] for type in a['b']['c']['d'].keys()]
Out[15]: [['conf1', 'conf2', 'conf3'], ['answer1', 'answer2', 'answer3']]

Upvotes: 4

jonrsharpe
jonrsharpe

Reputation: 122151

I would do that using recursive generator function:

def d_values(d, depth):
    if depth == 1:
        for i in d.values():
            yield i
    else:
        for v in d.values():
            if isinstance(v, dict):
                for i in d_values(v, depth-1):
                    yield i

Example:

>>> list(d_values({1: {2: 3, 4: 5}}, 2))
[3, 5]

In your case, this would give a dictionary like {'answer': answer1, 'conf': conf1} as each item, so you can use:

list(d['answer'] for d in d_values(a, 3))

Upvotes: 2

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