要列出或记录的元组的元组

Question

我正在编写一个job planner，下面是一个sqlite3查询结果。我想要的是按照工程师（例如“Dan”）和他们是否有工作（最后一列）或正在度假（第二列到最后一列）对数据进行分组

因此，理想情况下，我可能需要一个dict，每个工程师都有一把钥匙，每天都有一把钥匙，并且：

• 当天的工作清单
• 他们正在度假
• 他们没有工作 那天

我试过以下方法

jobs = {}
for j in res:
    jobs.setdefault(j[0], []).append(j[1:])


jobs2 = {}
for j in res:
    jobs2.setdefault((j[0], j[1]), []).append(j[2:])

但我不知道如何使用setdefault或dict理解来获得所需的数据结构。任何想法都值得赞赏

以下是数据：

('Dan', '2021-03-11', 'Thu', '2021-03-11', 'am', '11/03', 'Customer 1', '', 'U6kfoP9QPDw', None, 0, 1)                                                                                                                                 
('Dan', '2021-03-12', 'Fri', None, None, None, None, None, None, None, 0, 0)                                                                                                                                                                  
('Dan', '2021-03-13', 'Sat', None, None, None, None, None, None, None, 0, 0)                                                                                                                                                                  
('Dan', '2021-03-14', 'Sun', None, None, None, None, None, None, None, 0, 0)                                                                                                                                                                  
('Dan', '2021-03-15', 'Mon', None, None, None, None, None, None, None, 0, 0)                                                                                                                                                                  
('Dan', '2021-03-16', 'Tue', None, None, None, None, None, None, None, 0, 0)                                                                                                                                                                  
('Dan', '2021-03-17', 'Wed', None, None, None, None, None, None, None, 0, 0)                                                                                                                                                                  
('Gareth', '2021-03-11', 'Thu', None, None, None, None, None, None, None, 0, 0)                                                                                                                                                               
('Gareth', '2021-03-12', 'Fri', '2021-03-12', 'am', '12/03', 'Customer 4', '', 'k-uFnkwLLdo', None, 0, 2)                                                                                                                              
('Gareth', '2021-03-12', 'Fri', '2021-03-12', 'pm', '12/03', 'Customer 2', '', 'TWQdiG3piAE', None, 0, 2)
('Gareth', '2021-03-13', 'Sat', None, None, None, None, None, None, None, 0, 0)                                        
('Gareth', '2021-03-14', 'Sun', None, None, None, None, None, None, None, 0, 0)                                        
('Gareth', '2021-03-15', 'Mon', None, None, None, None, None, None, None, 0, 0)                                        
('Gareth', '2021-03-16', 'Tue', None, None, None, None, None, None, None, 0, 0)                                        
('Gareth', '2021-03-17', 'Wed', None, None, None, None, None, None, None, 0, 0)                                        
('Garth', '2021-03-11', 'Thu', None, None, None, None, None, None, None, 0, 0)                                         
('Garth', '2021-03-12', 'Fri', None, None, None, None, None, None, None, 0, 0)                                         
('Garth', '2021-03-13', 'Sat', None, None, None, None, None, None, None, 0, 0)                                         
('Garth', '2021-03-14', 'Sun', None, None, None, None, None, None, None, 0, 0)                                         
('Garth', '2021-03-15', 'Mon', None, None, None, None, None, None, None, 0, 0)                                         
('Garth', '2021-03-16', 'Tue', None, None, None, None, None, None, None, 0, 0)                                         
('Garth', '2021-03-17', 'Wed', None, None, None, None, None, None, None, 0, 0)                                         
('Ian', '2021-03-11', 'Thu', '2021-03-11', 'am', '11/03', 'Customer 3', '', 'SCfbuLeQ9ig', None, 0, 1)
('Ian', '2021-03-12', 'Fri', None, None, None, None, None, None, None, 0, 0)                                           
('Ian', '2021-03-13', 'Sat', None, None, None, None, None, None, None, 0, 0)                                           
('Ian', '2021-03-14', 'Sun', None, None, None, None, None, None, None, 0, 0)                                           
('Ian', '2021-03-15', 'Mon', None, None, None, None, None, None, None, 0, 0)                                           
('Ian', '2021-03-16', 'Tue', None, None, None, None, None, None, None, 0, 0)                                           
('Ian', '2021-03-17', 'Wed', None, None, None, None, None, None, None, 0, 0)                                           
('Jim', '2021-03-11', 'Thu', None, None, None, None, None, None, None, 0, 0)                                           
('Jim', '2021-03-12', 'Fri', None, None, None, None, None, None, None, 0, 0)                                           
('Jim', '2021-03-13', 'Sat', None, None, None, None, None, None, None, 0, 0)                                           
('Jim', '2021-03-14', 'Sun', None, None, None, None, None, None, None, 0, 0)                                           
('Jim', '2021-03-15', 'Mon', None, None, None, None, None, None, None, 0, 0)                                           
('Jim', '2021-03-16', 'Tue', None, None, None, None, None, None, None, 0, 0)                                           
('Jim', '2021-03-17', 'Wed', None, None, None, None, None, None, None, 1, 0)                    

Answer 1

看来一份口述就足够了。外部dict的键为员工姓名和值，内部dict的键为日期，值为作业/状态

ON_HOLIDAY = None

jobs = {}
for employee, date, *job in res:
    if employee not in jobs:
        jobs[employee] = dict()
    if date not in jobs[employee]:
        jobs[employee][date] = []
    if job[-2]:
        jobs[employee][date] = ON_HOLIDAY
    else:
        if job[-1]:
            jobs[employee][date].append(job[2:-2])

空列表表示该员工在该日期没有工作，但您也可以在解析完所有数据后使用另一个标识符更改该列表

Answer 2

多亏了Reti43的提示，我现在得到了以下结论：

jobs3 = {}
for nick, date, *job in res:
    jobs3.setdefault(nick, {}).setdefault(date, {}).setdefault('jobs', [])
    if job[-2]:
        jobs3[nick][date].update({'holiday': job[-3]})
    if job[-1]:
        jobs3[nick][date]['jobs'].append(job[2:-2])