Python 要列出或记录的元组的元组
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:Python 要列出或记录的元组的元组,python,sqlite,Python,Sqlite,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 =
- 当天的工作清单
- 他们正在度假
- 他们没有工作 那天
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)
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来更改该列表。多亏了Reti43的提示,我现在得出以下结论:
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])
那么您想要的确切输出格式是什么?你想要两本不同的字典吗?这就是你想要做的吗?如果你有更多的查找/分析要做,最好把你的数据放在一个数据框中。只有一个字典,每个工程师有一个键,每天有一个键。