从三个单独的列表创建了一个嵌套字典 [英] Created a nested dictionary from three separate lists

问题描述

我写了一些代码，这些代码生成了三个围绕项目的独立列表.第一个列表是ID列表，第二个列表是列表开始日期，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的字典中.

I wrote some code that generates three separate lists that revolve around projects. The first list is a list of IDs, the second list is a list start dates and third list is a list of end dates. I would like to combine these lists into a single nested dictionary.

item[0]和end_date中的item[0]与project_id[0]中的item[0]相关联

item[0] from start_date and item[0] from end_date are associated with 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)]

所需的输出:

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},
}

推荐答案

使用dict理解和zip:

Use a dict comprehension and 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)}}

如果要完整排序的字典，请使用以下命令:

If you want a completely ordered dict use one:

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))])}

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