包含多个字典列表的嵌套字典

Question

I have nested dictionary with lists like this

    {
     'a': 1,
     'x':[
          {'b': 1,
           'c': [
             {'z': 12},
             {'z': 22},
            ]
          },
          {'b': 2,
           'c': [
              {'z': 10},
              {'z': 33},
            ]
          }
        ]
     }

And I want to convert it to list of flat dictionaries i form like this.

[
 {'a': 1, 'b': 1, 'z': 12},
 {'a': 1, 'b': 1, 'z': 22},
 {'a': 1, 'b': 2, 'z': 10},
 {'a': 1, 'b': 2, 'z': 33},
]

Any idea how to achieve that?

Answer 1

A possible solution is:

#!/usr/bin/env python3
d = {
    'a': 1,
    'x': [
        {
            'b': 1,
            'c': [
                {'z': 12}
            ]
        },
        {
            'b': 2,
            'c': [
                {'z': 10}
            ]
        }
    ]
}

res = [{"a": 1, "b": x["b"], "z": x["c"][0]["z"]} for x in d["x"]]
print(res)

This assumes that there is only one a (with a fixed value of 1 ) and x element and this element is added to the comprehension manually.

The other two elements ( b and z ) are taken from x array with a list comprehension.

To learn more about how comprehensions work read the following:

• Python Documentation - 5.1.4. List Comprehensions
• Python: List Comprehensions

PS. You are supposed to first show what you have tried so far and get help on that. Take a look at SO rules before posting your next question.

Answer 2

The following produces the requested result:

[{'a': 1, 'b': 1, 'z': 12}, {'a': 1, 'b': 2, 'z': 10}]

Use at your own risk. The following was only tested on your example.

from itertools import product

def flatten(D):

    if not isinstance(D, dict): return D

    base = [(k, v) for k, v in D.items() if not isinstance(v, list)]
    lists = [[flatten(x) for x in v] for k, v in D.items() if isinstance(v, list)]

    l = []
    for p in product(*lists):
        r = dict(base)
        for a in p:
            for d in a:
                r.update(d)
        l.append(r)

    return l

The following tests above.

d = {
     'a': 1,
     'x':[
          {'b': 1,
           'c': [
             {'z': 12}
            ]
          },
          {'b': 2,
           'c': [
              {'z': 10}
            ]
          }
        ]
     }

print flatten(d)