Python Django:组合两个for循环
有没有办法将两个for循环组合成一个for循环? 我一直在寻找这样的解决方案。但事实并非如此:Python Django:组合两个for循环,python,django,Python,Django,有没有办法将两个for循环组合成一个for循环? 我一直在寻找这样的解决方案。但事实并非如此: ticket_stats_by_event = { x['pk']: {'sold_tickets': x['sold_tickets'], 'available_tickets: y['avaiable_tickets']} for x, y in sold_tickets_by_event, available_tickets_by_event } 结果将是: <Event
ticket_stats_by_event = {
x['pk']: {'sold_tickets': x['sold_tickets'], 'available_tickets: y['avaiable_tickets']}
for x, y in sold_tickets_by_event, available_tickets_by_event
}
结果将是:
<EventQuerySet [{'pk': 6, 'organizer': 1, 'available_tickets': 20, 'sold_tickets': 2}, {'pk': 1, 'organizer': 1, 'available_tickets': 1765746, 'sold_tickets': 2116}, {'pk': 5, 'organizer': 1, 'available_tickets': 10, 'sold_tickets': 1}, {'pk': 4, 'organizer': 1, 'available_tickets': 60, 'sold_tickets': 4}]>
以下是完整的功能:
def ticket_stats_by_event(self, organizer):
sold_tickets_by_event = [
x for x in self.sold_tickets if x['organizer'] == organizer.pk
]
# <EventQuerySet [{'pk': 6, 'organizer': 1, 'sold_tickets': 1}, {'pk': 1, 'organizer': 1, 'sold_tickets': 529}, {'pk': 5, 'organizer': 1, 'sold_tickets': 1}, {'pk': 4, 'organizer': 1, 'sold_tickets': 2}]>
available_tickets_by_event = [
x for x in self.available_tickets if x['organizer'] == organizer.pk
]
# <EventQuerySet [{'pk': 1, 'organizer': 1, 'available_tickets': 1721}, {'pk': 4, 'organizer': 1, 'available_tickets': 30}, {'pk': 5, 'organizer': 1, 'available_tickets': 10}, {'pk': 6, 'organizer': 1, 'available_tickets': 20}]>
ticket_stats_by_event = {
x['pk']: {'sold_tickets': x['sold_tickets']}
for x in sold_tickets_by_event
}
for item in available_tickets_by_event:
pk = item.pop('pk')
ticket_stats_by_event[pk].update(item)
return ticket_stats_by_event
def ticket\u stats\u by\u event(自我、组织者):
按活动售出的门票=[
如果x['organizer']==organizer.pk,则在self.self_门票中以x换x
]
#
可用的\u票证\u按\u事件=[
如果x['organizer']==organizer.pk,则在自助可用的\u票证中x代表x
]
#
票证统计按事件={
x['pk']:{'selled_tickets':x['selled_tickets']}
在每个活动中售出的x张门票
}
对于每个活动的可用门票中的项目:
pk=项目.pop('pk')
按事件[pk]记录的票证统计信息。更新(项目)
回程车票按事件统计
您可以只使用一个循环而不是四个循环来编写这样的函数:
def ticket_stats_by_event(self, organizer):
ticket_stats_by_event = {}
for st, at in zip(self.sold_tickets, self.available_tickets):
# if organizer is not same in any then skip
if organizer.pk != st['organizer'] or organizer.pk != at['organizer']:
continue
st_pk = st['pk']
st_value = {'sold_tickets': st['sold_tickets'], 'organizer': st['organizer']}
at_pk = at['pk']
at_value = {'available_tickets': at['available_tickets'], 'organizer': at['organizer']}
# add values from sold_tickets_by_event
if st_pk in ticket_stats_by_event:
ticket_stats_by_event[st_pk].update(st_value)
else:
ticket_stats_by_event[st_pk] = st_value
# add values from available_tickets_by_event
if at_pk in ticket_stats_by_event:
ticket_stats_by_event[at_pk].update(at_value)
else:
ticket_stats_by_event[at_pk] = at_value
return ticket_stats_by_event
此外,您可以将其分为三个循环,在第一个循环和第二个循环中,您可以获取特定组织的已售出和可用门票。我没有使用一行函数使其更具可读性。请展示您的模型并描述您试图实现的目标,而不是演示您如何做某事。