从三个单独的列表创建嵌套字典

我编写了一些代码来生成三个围绕项目的单独列表。第一个列表是 ID 列表，第二个列表是开始日期列表，第三个列表是结束日期列表。我想将这些列表合并到一个嵌套字典中。

item[0] from start_date and item[0]从 end_date 开始与item[0] from project_id[0]

project_id = ['project 1','project 2', 'project 3', 'project 4']

start_date = [datetime(2015,1,12), datetime(2015,1,13), datetime(2015,1,11), datetime(2015,1,13)]

end_date = [datetime(2015,1,15), datetime(2015,1,17), datetime(2015,1,15), datetime(2015,1,14)]

期望的输出：

d = {
    'project 1' : {'start date' : datetime(2015,1,12), 'end date' :  datetime(2015,1,15},
    'project 2' : {'start date' : datetime(2015,1,13), 'end date' :  datetime(2015,1,17},
    'project 3' : {'start date' : datetime(2015,1,11), 'end date' :  datetime(2015,1,15},
    'project 4' : {'start date' : datetime(2015,1,13), 'end date' :  datetime(2015,1,14},
}

使用 dict 理解和 zip：

d = {a:{"start date":b,"end date":c } for a,b,c  in zip(project_id,start_date,end_date)}

print d

{'project 4': {'end date': datetime.datetime(2015, 1, 14, 0, 0), 'start date': datetime.datetime(2015, 1, 13, 0, 0)}, 'project 2': {'end date': datetime.datetime(2015, 1, 17, 0, 0), 'start date': datetime.datetime(2015, 1, 13, 0, 0)}, 'project 3': {'end date': datetime.datetime(2015, 1, 15, 0, 0), 'start date': datetime.datetime(2015, 1, 11, 0, 0)}, 'project 1': {'end date': datetime.datetime(2015, 1, 15, 0, 0), 'start date': datetime.datetime(2015, 1, 12, 0, 0)}}

如果你想要一个完全有序的字典，请使用：

from collections import OrderedDict
d = OrderedDict()
for a,b,c in zip(project_id,start_date,end_date):
    d.setdefault(a,OrderedDict())
    d[a]["start date"] = b
    d[a]["end date"] = c
print(d)

{'project 1': OrderedDict([('start date', datetime.datetime(2015, 1, 12, 0, 0)), ('end date', datetime.datetime(2015, 1, 15, 0, 0))]),
'project 2': OrderedDict([('start date', datetime.datetime(2015, 1, 13, 0, 0)), ('end date', datetime.datetime(2015, 1, 17, 0, 0))]),
'project 3': OrderedDict([('start date', datetime.datetime(2015, 1, 11, 0, 0)), ('end date', datetime.datetime(2015, 1, 15, 0, 0))]),
'project 4': OrderedDict([('start date', datetime.datetime(2015, 1, 13, 0, 0)), ('end date', datetime.datetime(2015, 1, 14, 0, 0))])}