嵌套列表到嵌套字典

Question

我有一个包含主键的列表和一个列表列表的列表，其中每个封闭列表的第一个值（如 'key_01'）应是相应值的子键（如 'val_01', 'val_02').数据显示在这里：

master_keys = ["Master_01", "Master_02", "Master_03"]
data_long = [[['key_01','val_01','val_02'],['key_02','val_03','val_04'], ['key_03','val_05','val_06']],
           [['key_04','val_07','val_08'], ['key_05','val_09','val_10'], ['key_06','val_11','val_12']],
           [['key_07','val_13','val_14'], ['key_08','val_15','val_16'], ['key_09','val_17','val_18']]]

我希望将这些列表组合成字典的字典，如下所示：

master_dic = {
"Master_01": {'key_01':['val_01','val_02'],'key_02': ['val_03','val_04'], 'key_03': ['val_05','val_06']}, 
"Master_02": {'key_04': ['val_07','val_08'], 'key_05': ['val_09','val_10'], 'key_06': ['val_11','val_12']}, 
"Master_03": {'key_07': ['val_13','val_14'], ['key_08': ['val_15','val_16'], 'key_09': ['val_17','val_18']}
}

到目前为止我得到的是子字典：

import itertools

master_dic = {}
servant_dic = {}
keys = []
values = []
for line in data_long:
    for item in line:
        keys.extend(item[:1])
        values.append(item[1:])
servant_dic = dict(itertools.izip(keys, values))

正如预期的那样，它推出了一本字典。

servant_dic = {
'key_06': ['val_11','val_12'], 'key_04': ['val_08','val_07'], 'key_05': ['val_09','val_10'], 
'key_02': ['val_03','val_04'], 'key_03': ['val_05','val_06'], 'key_01': ['val_01','val_02']
 }

问题是，如果我想将 master_keys 添加到这个字典中，以便得到想要的结果，我必须按特定顺序执行此操作，这是可能的，如果每行都有一个这样的计数器：

enumerated_dic =
{
0: {'key_01':['val_01','val_02'],'key_02': ['val_03','val_04'], 'key_03': ['val_05','val_06']}, 
1: {'key_04': ['val_07','val_08'], 'key_05': ['val_09','val_10'], 'key_06': ['val_11','val_12']}, 
2: {'key_07': ['val_13','val_14'], ['key_08': ['val_15','val_16'], 'key_09': ['val_17','val_18']}
}

我很想用 enumerate() 来做这个，而 servant_dic 的每一行都是构建的，但不知道如何构建。从那以后，我可以简单地将计数器 0、1、2 等替换为 master_keys。

感谢您的帮助。

Answer 1

master_keys = ["Master_01", "Master_02", "Master_03"]
data_long = [[['key_01','val_01','val_02'],['key_02','val_03','val_04'], ['key_03','val_05','val_06']],
           [['key_04','val_07','val_08'], ['key_05','val_09','val_10'], ['key_06','val_11','val_12']],
           [['key_07','val_13','val_14'], ['key_08','val_15','val_16'], ['key_09','val_17','val_18']]]


_dict = {}

for master_key, item in zip(master_keys, data_long):
    _dict[master_key] = {x[0]: x[1:] for x in item}

print _dict

Answer 2

您还可以使用 pop 和字典理解：

for key, elements in zip(master_keys, data_long):
    print {key: {el.pop(0): el for el in elements}}
   ...:
{'Master_01': {'key_02': ['val_03', 'val_04'], 'key_03': ['val_05', 'val_06']}}
{'Master_02': {'key_06': ['val_11', 'val_12'], 'key_04': ['val_07', 'val_08'], 'key_05': ['val_09', 'val_10']}}
{'Master_03': {'key_07': ['val_13', 'val_14'], 'key_08': ['val_15', 'val_16'], 'key_09': ['val_17', 'val_18']}}

Answer 3

希望这会有所帮助：

{master_key: {i[0]: i[1:] for i in subkeys} for master_key, subkeys in zip(master_keys, data_long)}

Answer 4

我的功能方法：

master_dic = dict(zip(master_keys, [{k[0]: k[1::] for k in emb_list} for emb_list in data_long]))
print(master_dic)

嵌套列表到嵌套字典

Nested lists to nested dicts

python

list

dictionary

enumerate