Python如何根据日期创建子列表？

Question

I'm have a list like this:

my = [datetime.datetime(2013, 11, 12, 20, 22, 5),
     datetime.datetime(2013, 11, 12, 20, 21, 27),
     datetime.datetime(2013, 11, 12, 20, 20, 28),
     datetime.datetime(2013, 11, 11, 18, 21, 27),
     datetime.datetime(2013, 11, 11, 15, 21, 27),
     datetime.datetime(2013, 11, 11, 11, 21, 27),
     datetime.datetime(2013, 11, 10, 9, 21, 27),
     datetime.datetime(2013, 11, 10, 19, 21, 27),
     datetime.datetime(2013, 11, 10, 22, 21, 27),
     datetime.datetime(2013, 11, 10, 20, 21, 27)]

And I want create lists by day into another list, I'm using a[n].date() to compare dates and join them, but I'm running in circles whithout get it, please could someone help me?

Result expected:

    days = [
    [datetime.datetime(2013, 11, 12, 20, 22, 5), datetime.datetime(2013, 11, 12, 20, 21, 27), datetime.datetime(2013, 11, 12, 20, 20, 28)],
    [datetime.datetime(2013, 11, 11, 18, 21, 27),datetime.datetime(2013, 11, 11, 15, 21, 27),datetime.datetime(2013, 11, 11, 11, 21, 27)],
    [datetime.datetime(2013, 11, 10, 9, 21, 27),datetime.datetime(2013, 11, 10, 19, 21, 27),datetime.datetime(2013, 11, 10, 22, 21, 27),datetime.datetime(2013, 11, 10, 20, 21, 27)]
           ]

Attempt failed:

In [152]: fecha = None
   .....: dia = []
   .....: days = []
   .....: for i in my:
   .....:     if fecha:
   .....:         if i.date() == fecha:
   .....:             dia.append(i)
   .....:         else:
   .....:             days.append(dia)
   .....:             fecha = i.date()
   .....:     else:
   .....:         fecha = i.date()

Thanks

Answer 1

Use itertools.groupby and group using key = datetime.date() :

.date() on a datetime object return date object, we can use this to group the items of the list:

>>> my[0].date()
datetime.date(2013, 11, 12)

Demo:

>>> from pprint import pprint
>>> from itertools import groupby
>>> pprint([list(g) for k, g in groupby(my, key=lambda x:x.date())])
[[datetime.datetime(2013, 11, 12, 20, 22, 5),
  datetime.datetime(2013, 11, 12, 20, 21, 27),
  datetime.datetime(2013, 11, 12, 20, 20, 28)],
 [datetime.datetime(2013, 11, 11, 18, 21, 27),
  datetime.datetime(2013, 11, 11, 15, 21, 27),
  datetime.datetime(2013, 11, 11, 11, 21, 27)],
 [datetime.datetime(2013, 11, 10, 9, 21, 27),
  datetime.datetime(2013, 11, 10, 19, 21, 27),
  datetime.datetime(2013, 11, 10, 22, 21, 27),
  datetime.datetime(2013, 11, 10, 20, 21, 27)]]

If keys are not already sorted by dates then use a collections.defaultdict(list) and use datetime.date() as key.

Answer 2

Okay Goku, let me show you how to use itertools.groupby . Here:

for date, group in itertools.groupby(my, lambda x: x.day):
    new_list.append(list(group))

Or list comprehension:

new_list = [list(group) for day, group in itertools.groupby(my, lambda x: x.day)]

Demo:

>>> import itertools
>>> new_list = []
>>> for date, group in itertools.groupby(my, lambda x: x.day):
    new_list.append(list(group))


>>> new_list
[[datetime.datetime(2013, 11, 12, 20, 22, 5), datetime.datetime(2013, 11, 12, 20, 21, 27), datetime.datetime(2013, 11, 12, 20, 20, 28)], [datetime.datetime(2013, 11, 11, 18, 21, 27), datetime.datetime(2013, 11, 11, 15, 21, 27), datetime.datetime(2013, 11, 11, 11, 21, 27)], [datetime.datetime(2013, 11, 10, 9, 21, 27), datetime.datetime(2013, 11, 10, 19, 21, 27), datetime.datetime(2013, 11, 10, 22, 21, 27), datetime.datetime(2013, 11, 10, 20, 21, 27)]]
>>> 

Edit: Sorry, i thought you were grouping it by day. If you want to group it by date, do this:

for date, group in itertools.groupby(my, lambda x: x.date()):
    new_list.append(list(group))

Good luck!