子集的元組密鑰詞典與名字典基於鍵

Question

元組鍵（KEY0，KEY1）字典（df_dict）我想子集中有幾個dataframes，每個數據幀中的列日期和accountNum。我想要子集df_dict並基於key0生成字典名稱。

df_dict = {('100', '001'): date, accountNum, ('100', '002'): date, accountNum, 
      ('200', '001'): date, accountNum, ('200', '002'): date, accountNum} 

在df_dict的dataframes如下所示，

('100','001')-DataFrame ('100','002')-DataFrame ('200','001')-DataFrame 
date  accountNum date  accountNum data  accountNum 
2010-01-01  280  2010-02-01  150  2010-03-01  330 
2010-01-02  285  2010-02-02  155  2010-03-02  335 
2010-01-03  290  2010-02-03  160  2010-03-03  340 

('200','002')-DataFrame 
date  accountNum 
2010-04-01  510 
2010-04-02  515 
2010-04-03  520 

我預計會是這樣的結果，

df_dict_100 = {('100', '001'): date, accountNum, ('100','002'): date, accountNum} 
df_dict_200 = {('200', '001'): date, accountNum, ('200','002'): date, accountNum} 

而且在每個字典中的數據幀將像，

df_dict100 
('100','001')-DataFrame ('100','002')-DataFrame 
date  accountNum date  accountNum 
2010-01-01  280  2010-02-01  150  
2010-01-02  285  2010-02-02  155  
2010-01-03  290  2010-02-03  160  

df_dict200 
('200','001')-DataFrame ('200','002')-DataFrame 
date  accountNum date   accountNum 
2010-01-01  280  2010-04-01  510 
2010-01-02  285  2010-04-02  515 
2010-01-03  290  2010-04-03  520 

這個是我的做法，

my_list = ['100','200'] 
subset_dict = {k: v for k, v in df_dict.items() if k[0] in my_list} 

但似乎我從df_dict得到確切的字典。

Answer 1

通過創建多級字典，可以將第一種形式轉換爲第二種形式。因此，而不是df_dict_100，你可能有df_dict[100]，就像這樣：

import pprint 

date, accountNum = 'date', 'accountNum' 
df_dict = {('100', '001'): (date, accountNum), ('100', '002'): (date, accountNum), 
      ('200', '001'): (date, accountNum), ('200', '002'): (date, accountNum)} 

new_dict = dict() 
for key, value in df_dict.items(): 
    new_dict.setdefault(key[0], {})[key] = value 

pprint.pprint(new_dict) 

所以，結果是：

{'100': {('100', '001'): ('date', 'accountNum'), 
     ('100', '002'): ('date', 'accountNum')}, 
'200': {('200', '001'): ('date', 'accountNum'), 
     ('200', '002'): ('date', 'accountNum')}} 

要訪問個人數據，你可能會使用類似語法：

print(new_dict['100']['100', '001'][0]) 

如果你喜歡字典理解，試試這個：

subset_dict = { 
    matching_key : { 
     k: v for k, v in df_dict.items() if k[0] == matching_key } 
    for matching_key in set(k[0] for k in df_dict) 
} 

在評論中，OP詢問「我可以知道如何生成兩個字典，而不是一個字典中的兩個字典？」像這樣的東西應該工作：

df_dict_100 = { k: v for k, v in df_dict.items() if k[0] == '100' } 
df_dict_200 = { k: v for k, v in df_dict.items() if k[0] == '200' } 

把這些彙集成for環路，這裏是一個完整的程序：

import pprint 

date, accountNum = 'date', 'accountNum' 
df_dict = {('100', '001'): (date, accountNum), ('100', '002'): (date, accountNum), 
      ('200', '001'): (date, accountNum), ('200', '002'): (date, accountNum)} 

my_list = ['100', '200'] 
for i in my_list: 
    new_df_dict = { k: v for k, v in df_dict.items() if k[0] == i } 
    pprint.pprint(new_df_dict) 
    print("----") 

這裏是輸出：

{('100', '001'): ('date', 'accountNum'), 
('100', '002'): ('date', 'accountNum')} 
---- 
{('200', '001'): ('date', 'accountNum'), 
('200', '002'): ('date', 'accountNum')} 
----