Python字典问题：按元组键中的元素分组

Question

所以我有一个类似这样的字典，有4个元素元组作为键，列表列表作为相应的值。（yay索引）

{('A002', 'R051', '02-00-00', 'LEXINGTON AVE'): [[datetime.datetime(2015, 6, 20, 0, 0),
                                                  750],
                                                 [datetime.datetime(2015, 6, 21, 0, 0),
                                                  576],
                                                 [datetime.datetime(2015, 6, 22, 0, 0),
                                                  1486],
                                                 [datetime.datetime(2015, 6, 23, 0, 0),
                                                  595],
                                                 [datetime.datetime(2015, 6, 24, 0, 0),
                                                  841],
                                                 [datetime.datetime(2015, 6, 25, 0, 0),
                                                  1072],
                                                 [datetime.datetime(2015, 6, 26, 0, 0),
                                                  1049]],
 ('A002', 'R051', '02-00-01', 'LEXINGTON AVE'): [[datetime.datetime(2015, 6, 20, 0, 0),
                                                  670],
                                                 [datetime.datetime(2015, 6, 21, 0, 0),
                                                  457],
                                                 [datetime.datetime(2015, 6, 22, 0, 0),
                                                  1189],
                                                 [datetime.datetime(2015, 6, 23, 0, 0),
                                                  505],
                                                 [datetime.datetime(2015, 6, 24, 0, 0),
                                                  665],
                                                 [datetime.datetime(2015, 6, 25, 0, 0),
                                                  354],
                                                 [datetime.datetime(2015, 6, 26, 0, 0),
                                                  651]]}

我想修改这个字典，以便合并具有相同的第一、第二和第四元组元素的所有键的值。（就像上面的两把钥匙一样）。我想将这两个键元组组合成一个键元组（这样我的组合键就是('A002', 'R051', 'LEXINGTON AVE')），并合并这些值。在python中这是可能的吗？在

例如，第一个值是[日期时间。日期时间（2015，6，20，0，0），1420]——本例中为670+750

提前谢谢。在

Answer 1

是的，从Python2.7以后使用groupby和{}是非常有可能的。在

示例代码-

>>> from itertools import groupby
>>> import datetime
>>> d = {('A002', 'R051', '02-00-00', 'LEXINGTON AVE'): [[datetime.datetime(2015, 6, 20, 0, 0),
...                                                   750],
...                                                  [datetime.datetime(2015, 6, 21, 0, 0),
...                                                   576],
...                                                  [datetime.datetime(2015, 6, 22, 0, 0),
...                                                   1486],
...                                                  [datetime.datetime(2015, 6, 23, 0, 0),
...                                                   595],
...                                                  [datetime.datetime(2015, 6, 24, 0, 0),
...                                                   841],
...                                                  [datetime.datetime(2015, 6, 25, 0, 0),
...                                                   1072],
...                                                  [datetime.datetime(2015, 6, 26, 0, 0),
...                                                   1049]],
...  ('A002', 'R051', '02-00-01', 'LEXINGTON AVE'): [[datetime.datetime(2015, 6, 20, 0, 0),
...                                                   670],
...                                                  [datetime.datetime(2015, 6, 21, 0, 0),
...                                                   457],
...                                                  [datetime.datetime(2015, 6, 22, 0, 0),
...                                                   1189],
...                                                  [datetime.datetime(2015, 6, 23, 0, 0),
...                                                   505],
...                                                  [datetime.datetime(2015, 6, 24, 0, 0),
...                                                   665],
...                                                  [datetime.datetime(2015, 6, 25, 0, 0),
...                                                   354],
...                                                  [datetime.datetime(2015, 6, 26, 0, 0),
...                                                   651]]}
>>>
>>> newd = {(x[0],x[1],x[2]):[z for a in y for z in a[1]] for x, y in groupby(d.items(),key= lambda x: (x[0][0],x[0][1],x[0][3]))}
>>> newd
{('A002', 'R051', 'LEXINGTON AVE'): [[datetime.datetime(2015, 6, 20, 0, 0), 750], [datetime.datetime(2015, 6, 21, 0, 0), 576], [datetime.datetime(2015, 6, 22, 0, 0), 1486], [datetime.datetime(2015, 6, 23, 0, 0), 595], [datetime.datetime(2015, 6, 24, 0, 0), 841], [datetime.datetime(2015, 6, 25, 0, 0), 1072], [datetime.datetime(2015, 6, 26, 0, 0), 1049], [datetime.datetime(2015, 6, 20, 0, 0), 670],
[datetime.datetime(2015, 6, 21, 0, 0), 457], [datetime.datetime(2015, 6, 22, 0, 0), 1189], [datetime.datetime(2015, 6, 23, 0, 0), 505], [datetime.datetime(2015, 6, 24, 0, 0), 665], [datetime.datetime(2015, 6, 25, 0, 0), 354], [datetime.datetime(2015, 6, 26, 0, 0), 651]]}