将也有列表的嵌套字典展平为字典

Question

我有一个嵌套字典，里面有一个列表，例如

{'Ka': {'Ka': ['0.80', '0.1'],
  'Ba': ['0.50', '1.1'],
  'FC': ['0.78', '0.0'],
  'AA': ['0.66', '8.1']},
 'AL': {'AR': ['2.71', '7.3'], 'KK': ['10.00', '90.0']}}

我想将其拼合成字典

{'Ka': {'Ka.0':'0.80', 'Ka.1':'0.1',
  'Ba.0':'0.50','Ba.1': '1.1',
  'FC.0':'0.78', 'FC.1':'0.0',
  'AA.0':'0.66', 'AA.1':'8.1'},
 'AL': {'AR.0':'2.71','AR.1':'7.3', 'KK.0':'10.00','KK.1':'90.0'}}

我尝试使用以下方法来展平这个嵌套字典：

import collections
def flatten(dictionary, parent_key=False, separator='.'):
    """
    Turn a nested dictionary into a flattened dictionary
    :param dictionary: The dictionary to flatten
    :param parent_key: The string to prepend to dictionary's keys
    :param separator: The string used to separate flattened keys
    :return: A flattened dictionary
    """

    items = []
    for key, value in dictionary.items():
        new_key = str(parent_key) + separator + key if parent_key else key
        if isinstance(value, collections.MutableMapping):
            items.extend(flatten(value, new_key, separator).items())
        elif isinstance(value, list):
            for k, v in enumerate(value):
                items.extend(flatten({str(k): v}, new_key).items())
        else:
            items.append((new_key, value))
    return dict(items)

并获得

{'Ka.Ka.0': '0.80',
 'Ka.Ka.1': '0.1',
 'Ka.Ba.0': '0.50',
 'Ka.Ba.1': '1.1',
 'Ka.FC.0': '0.78',
 'Ka.FC.1': '0.0',
 'Ka.AA.0': '0.66',
 'Ka.AA.1': '8.1',
 'AL.AR.0': '2.71',
 'AL.AR.1': '7.3',
 'AL.KK.0': '10.00',
 'AL.KK.1': '90.0'}

但这不是我想要的。我想得到看起来像上面第一个的东西。提前致谢。

Answer 1

# use a nested dict comprehension
# use enumerate for the index of the list items and add it to the key using f-string
{key: {f"{k}.{i}": e for k, v in val.items() for i, e in enumerate(v)} for key, val in my_dict.items()}
{'Ka': {'Ka.0': '0.80', 'Ka.1': '0.1',
        'Ba.0': '0.50', 'Ba.1': '1.1',
        'FC.0': '0.78', 'FC.1': '0.0',
        'AA.0': '0.66', 'AA.1': '8.1'},
 'AL': {'AR.0': '2.71', 'AR.1': '7.3', 
        'KK.0': '10.00', 'KK.1': '90.0'}}

Answer 2

sample_dict = {'Ka': {'Ka': ['0.80', '0.1'],
  'Ba': ['0.50', '1.1'],
  'FC': ['0.78', '0.0'],
  'AA': ['0.66', '8.1']},
 'AL': {'AR': ['2.71', '7.3'], 'KK': ['10.00', '90.0']}}
 
for item, value in sample_dict.items():
    print(item, value, type(value))
    if type(value)==dict:
        for item2 in list(value):
            if type(value[item2])==list:
                value[item2 + ".0"] = value[item2][0]
                value[item2 + ".1"] = value[item2][1]
                del value[item2]
print(sample_dict)

结果是：

{'Ka': {'Ka.0': '0.80', 'Ka.1': '0.1', 
        'Ba.0': '0.50', 'Ba.1': '1.1', 
        'FC.0': '0.78', 'FC.1': '0.0', 
        'AA.0': '0.66', 'AA.1': '8.1'}, 
 'AL': {'AR.0': '2.71', 'AR.1': '7.3', 
        'KK.0': '10.00', 'KK.1': '90.0'}}

Answer 3

from collections import defaultdict
new = defaultdict(dict)
for k, values in d.items():
    for sub_key, values in values.items():
        for value in values:
            existing_key_count = sum(1 for existing_key in new[k].keys() if existing_key.startswith(sub_key))
            new_key = f"{sub_key}.{existing_key_count}"
            new[k][new_key] = value

{'Ka': {'Ka.0': '0.80', 'Ka.1': '0.1', 'Ba.0': '0.50', 'Ba.1': '1.1', 'FC.0': '0.78', 'FC.1': '0.0', 'AA.0': '0.66', 'AA.1': '8.1'}, 'AL': {'AR.0': '2.71', 'AR.1': '7.3', 'KK.0': '10.00', 'KK.1': '90.0'}}

将也有列表的嵌套字典展平为字典

Flatten the nested dictionary, which has list also, into a dictionary

python

list

dictionary

nested

flatten