极富挑战性的查询集过滤，排序和注释（在基于Django的应用中）

Question

I have a Django-based web-app where users congregate and chat with one another. I've just finished writing a feature whereby users can make their own "chat groups", centered around any topic of interest. These could either be private , or publically visible.

My next challenge is showing a list of all existing public groups , paginated, and sorted by the most happening group first . After some deep thinking, I've decided that the most happening group is one which sees the most unique visitors (silent or otherwise) in the previous 60 mins . To be sure, by unique I mean distinct users, and not the same user hitting a group again and again.

What's the most efficient way to get the desired, ordered query-set in the get_queryset() method of my class-based view associated to this popularity listing? Secondly, what's the most efficient way to also annotate total distinct views to each group object in the same queryset, so that I can additionally show total views, while sort according to what's currently hot?

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Relevant models are:

class Group(models.Model):
    topic = models.TextField(validators=[MaxLengthValidator(200)], null=True)
    owner = models.ForeignKey(User)
    private = models.CharField(max_length=50, default=0)
    created_at = models.DateTimeField(auto_now_add=True)

class GroupTraffic(models.Model):
    visitor = models.ForeignKey(User)
    which_group = models.ForeignKey(Group)
    time = models.DateTimeField(auto_now_add=True)

The relevant view is:

class GroupListView(ListView):
    model = Group
    form_class = GroupListForm
    template_name = "group_list.html"
    paginate_by = 25

    def get_queryset(self):
        return Group.objects.filter(private=0,date__gte=???).distinct('grouptraffic__visitor').annotate(recent_views=Count('grouptraffic__???')).order_by('-recent_views').annotate(total_views=Count('grouptraffic__which_group=group'))

As you can see, I've struggled rather mightily in the get_queryset(self) method above, annotating twice and what not. Please advise!

Answer 1

You cannot combine annotate() and distinct() in a single django query. So you can try like:

date = datetime.datetime.now()-datetime.timedelta(hours=1)

Next query is to get the grouptraffic with unique visitors

new_traff = GroupTraffic.objects.filter(time__gte=date).distinct('visitor','which_group').values_list('id',flat=True)

trendingGrp_ids = GroupTraffic.objects.filter(id__in=new_traff).values('which_group').annotate(total=Count('which_group')).order_by('-total')

The above query will get you trending groupids ordered by total like:

[{'total': 4, 'which_group': 2}, {'total': 2, 'which_group': 1}, {'total': 1, 'which_group': 3}]

Here total refers to no. of new unique visitors for each group in the last 60 minutes.

Now iterate over trendingGrp_ids to get the trending trendingGrps with views:

trendingGrps = [Group.objects.filter(id=grp['which_group']).extra(select={"views":grp['total']})[0] for grp in trendingGrp_ids]

Update:

To get all public groups, and sort them by how hot they are via measuring the traffic they received in the past 1 hr.

new_traff = GroupTraffic.objects.filter(time__gte=date,which_group__private=0).distinct('visitor','which_group').values_list('id',flat=True)

trendingGrp_ids = GroupTraffic.objects.filter(id__in=new_traff).values('which_group').annotate(total=Count('which_group')).order_by('-total')

trendingGrps = [Group.objects.filter(id=grp['which_group']).extra(select={"views":grp['total']})[0] for grp in trendingGrp_ids]

trndids = [grp['which_group'] for grp in trendingGrp_ids]

nonTrendingGrps = Group.objects.filter(private=0).exclude(id__in=trndids).extra(select={"views":0})

allGrps = trendingGrps.append(nonTrendingGrps)

Answer 2

1)Create a separate function that tallies the distinct views in a chatbox. For every chatbox, put the result in a list. Return the biggest value in the list and assign it to a variable. Import the function and filter with the variable.

2) Make a set for each box. The set contains all the distinct users of the chatbox. Filter by the lenght of the set.