如何用相同的键连接字典的值

Question

我有一本字典，其中包含一些字典（其编号不固定），而字典又包含数字列表（每个子词典的列表数量也不固定）。

Data_Dict ={0:{0:[list 00], 1:[list 01], 2:[list 02], 3:[list 03], ...},
            1:{0:[list 10], 1:[list 11], 2:[list 12], 3:[list 13], ...},
            2:{0:[list 20], 1:[list 21], 2:[list 22], 3:[list 23], ...},
            ...}

Data_Dict 是我从不同文件中解析数据得到的，每个文件都提供有关不同对象的信息。

字典的键来自我用来导入不同文件或不同对象的for循环（更具体地说，当值是字典时，键是指文件，而当值是字典时，键是指单个对象values 是字符串）。

我要做的是将列表与单个列表中所有子词典的特定键连接起来。保持顺序很重要，这意味着列表 Data_Dict[0][0] 的元素必须位于列表 Data_Dict[1][0] 的元素之前，依此类推。在这里你有想要的结果：

NewData_Dict ={0:[list 00, list 10, list 20,...],
               1:[list 01, list 11, list 21,...],
               2:[list 02, list 12, list 22,...], 
               ...}

我一直在尝试使用另一个问题 (How can I combine dictionaries with the same keys?) 中的建议，即：

NewData_Dict = {}
for k in Data_Dict[0]:
    NewData_Dict[k] = [d[k] for d in Data_Dict]

它给我的是：

NewData_Dict[k] = [d[k] for d in Data_Dict]
TypeError: 'int' object is not subscriptable

我也尝试了问题Combine values of same keys in a list of dicts的答案：

NewData_Dict = {
    k: [d.get(k) for d in Data_Dict]
    for k in set().union(*Data_Dict)
}

得到回报：

NewData_Dict = {k: [d.get(k) for d in Data_Dict] for k in set().union(*Data_Dict)}
TypeError: 'int' object is not iterable

我认为问题可能是我的字典的键是“int”。我的假设正确吗？我怎样才能解决这个问题？

Answer 1

这个循环可能会有所帮助

the_new_dic = {}
for key_1 in range(len(the_dict)):
  for key_2 in range(len(the_dict[key_1])):
    the_new_dic[key_1].extend(the_dict[key_2][key_1])

Answer 2

您可以遍历每个嵌套字典的项目。

data = {0: {0: ['00', '00'], 1: ['01', '01'], 2: ['02', '02'], 3: ['03', '03']},
        1: {0: ['10', '10'], 1: ['11', '11'], 2: ['12', '12'], 3: ['13', '13']},
        2: {0: ['20', '20'], 1: ['21', '21'], 2: ['22', '22'], 3: ['23', '23']}}

new = {}
for x in data.values():
    for key in x:
        try:
            new[key].extend(x[key])
        except KeyError:
            new[key] = x[key]

输出：

>>> new
{0: ['00', '00', '10', '10', '20', '20'], 
 1: ['01', '01', '11', '11', '21', '21'], 
 2: ['02', '02', '12', '12', '22', '22'], 
 3: ['03', '03', '13', '13', '23', '23']}

Answer 3

for i in range(len(Data_dict)):
    for j in range(len(Data_dict[i] - 1):
        Data_dict[i][0].extend(Data_dict[i][j+1]

Answer 4

代码 cmets 中的解释。试试：

Data_Dict ={0:{0:["A", "B", "C"], 1:["E", "F", "G"], 2:["H", "I", "J"], 3:["K", "L", "M"]},
            1:{0:["N", "O", "P"], 1:["Q", "R", "S"], 2:["T", "U", "V"], 3:["W", "X", "Y"]},
            2:{0:["AA", "BB", "CC"], 1:["DD","EE","FF"], 2:["GG", "HH", "II"], 3:["JJ","KK","LL"]},}

New_Data_Dict = {}
# for each key in Data_Dict
for dict in Data_Dict:
    # for each key in Data_Dict[key]
    for list in Data_Dict[dict]:
        # if list key exists in New_Data_Dict append Data_Dict[dict key][list key] <<== Value of list
        if list in New_Data_Dict:
            New_Data_Dict[list]+=(Data_Dict[dict][list])
        # else create list key and add value
        else:
            New_Data_Dict[list] = Data_Dict[dict][list]

print(New_Data_Dict)

输出：

{0: ['A', 'B', 'C', 'N', 'O', 'P', 'AA', 'BB', 'CC'], 
 1: ['E', 'F', 'G', 'Q', 'R', 'S', 'DD', 'EE', 'FF'], 
 2: ['H', 'I', 'J', 'T', 'U', 'V', 'GG', 'HH', 'II'], 
 3: ['K', 'L', 'M', 'W', 'X', 'Y', 'JJ', 'KK', 'LL']}