如果条件为True,则装饰一个函数？

我不会为此编写装饰器 - 请注意,装饰器语法允许返回装饰器的函数调用(实际上,这是decorators-with-parameters实际执行的操作).所以:

def maybe_decorate(condition, decorator):
    return decorator if condition else lambda x: x

@api_view(['GET'])
@maybe_decorate(settings.USE_API_KEY, authentication_classes((TokenAuthentication, )))
@maybe_decorate(settings.USE_API_KEY, permission_classes((IsAuthenticated, )))
def all_announcements(request):
    ...

当你使用这样的装饰器时:

@decorator
def f():
    ...

它相当于:

def f():
    ...
f = decorator(f)

考虑到这一点,你可以完成你想要的东西:

def all_announcements(request):
    announcements = Announcement.objects.all()
    serializer = AnnouncementSerializer(announcements, many=True)
    return Response(serializer.data)

if settings.USE_API_KEY:
    all_announcements = permission_classes((IsAuthenticated,))(all_announcements)
    all_announcements = authentication_classes((TokenAuthentication,))(all_announcements)
all_announcements = api_view(['GET'])(all_announcements)