Django模型组按日期时间的日期 [英] Django model group by datetime's date
本文介绍了Django模型组按日期时间的日期的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
类实体(models.Model):
start_time = models。 DateTimeField()
我想将它们重新分组为列表列表列表列表包含同一日期的实体(同一天,时间应该被忽略)。
如何以pythonic方式实现?
谢谢
解决方案
创建一个小功能来提取日期:
def extract_date(entity):
'从实体'
中提取起始日期return entity.start_time.date()
然后你可以使用它与 itertools.groupby
:
from itertools import groupby
entities = Entity.objects.order_by('start_time')
for start_date,groupby (实体,key = extract_date):
do_something_with(start_date,list(group))
或者,如果您真的想要列表列表:
entities = Entity.objects.order_by('start_time')
list_of_lists = [list(g)for t,g in groupby(entities,key = extract_date)]
Assume I have a such model:
class Entity(models.Model):
start_time = models.DateTimeField()
I want to regroup them as list of lists which each list of lists contains Entities from the same date (same day, time should be ignored).
How can this be achieved in a pythonic way ?
Thanks
解决方案
Create a small function to extract just the date:
def extract_date(entity):
'extracts the starting date from an entity'
return entity.start_time.date()
Then you can use it with itertools.groupby
:
from itertools import groupby
entities = Entity.objects.order_by('start_time')
for start_date, group in groupby(entities, key=extract_date):
do_something_with(start_date, list(group))
Or, if you really want a list of lists:
entities = Entity.objects.order_by('start_time')
list_of_lists = [list(g) for t, g in groupby(entities, key=extract_date)]
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