Get all values from nested dictionaries in python

问题:

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()]

回答1:

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}}}
list(NestedDictValues(a))

The output is:

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

which is all of the values.

回答2:

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']]

回答3:

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))

回答4:

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.

回答5:

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