djangodjango-queryset

Anotation after filter (freeze queryset) Django


In the process of model annotation there is a need to filter the finished list. But the value of the annotation "rank" after the filter() becomes "1" because it has only one element. Without filtering the queryset everything works fine

request_user = (
            MainUser.objects.select_related("balance_account")
            .annotate(coin_balance=F("balance_account__coin_balance"))
            .annotate(rank=Window(expression=RowNumber(), order_by="-coin_balance"))
            .filter(id=data.get("user_id"))
            .first()
        )

Is there a way to avoid the race or freeze the filtered queryset?

qs = set(
      MainUser.objects.select_related("balance_account")
     .annotate(coin_balance=F("balance_account__coin_balance"))  
     .annotate(rank=Window(expression=RowNumber(), 
      order_by="-coin_balance"))
     )
q = list(filter(lambda x: x.id == data.get("user_id"), qs))[0]

But how optimal is this approach?


Solution

  • In your code, the issue arises because Django's Window functions, like RowNumber(), calculate the rank in the context of the entire queryset. When you apply filter(), it reduces the queryset to a single element, making the rank "1." To retain the original rank, you need to calculate it before filtering, and then filter the result without changing the annotated rank.

    Here are a few optimized approaches to address this issue:

    1. Use a Subquery for Pre-Filtered Ranking

    One approach is to compute the ranking in a subquery and then filter on the main queryset. This way, the ranking is preserved even after filtering. Here’s how you could implement it:

    from django.db.models import OuterRef, Subquery, Window, F
    from django.db.models.functions import RowNumber
    
    # Calculate the rank for each user and store it in a subquery
    subquery = MainUser.objects.select_related("balance_account").annotate(
        coin_balance=F("balance_account__coin_balance"),
        rank=Window(expression=RowNumber(), order_by=F("coin_balance").desc())
    ).filter(id=OuterRef("pk"))
    
    # Use the subquery to get the user with preserved rank
    request_user = MainUser.objects.annotate(
        coin_balance=F("balance_account__coin_balance"),
        rank=Subquery(subquery.values("rank")[:1])
    ).filter(id=data.get("user_id")).first()
    

    This approach ensures the rank annotation is computed across the entire set of users based on coin_balance before filtering.

    2. Rank in a Separate Query and Cache

    Another approach is to rank all users and cache the result, then filter for the specific user. This avoids recalculating the rank each time and ensures the rank remains consistent.

    # Cache all users with rank and balance
    ranked_users = list(
        MainUser.objects.select_related("balance_account")
        .annotate(coin_balance=F("balance_account__coin_balance"))
        .annotate(rank=Window(expression=RowNumber(), order_by=F("coin_balance").desc()))
    )
    
    # Filter the cached list for the user with the specified ID
    request_user = next((user for user in ranked_users if user.id == data.get("user_id")), None)
    

    This approach is optimal if you only need to look up users occasionally after ranking all of them. However, it may not be ideal for large datasets due to the memory usage of storing all users in a list.

    3. Use annotate(rank=Subquery(...)) with a Materialized List

    If your dataset is manageable in memory, you could first get a materialized list of users with ranks and then filter it:

    # Get all users with ranking in a materialized queryset
    all_users_ranked = list(
        MainUser.objects.select_related("balance_account")
        .annotate(coin_balance=F("balance_account__coin_balance"))
        .annotate(rank=Window(expression=RowNumber(), order_by=F("coin_balance").desc()))
    )
    
    # Filter to find the specific user by ID without affecting rank
    request_user = next((user for user in all_users_ranked if user.id == data.get("user_id")), None)
    

    Each of these solutions avoids re-ranking or losing rank information and improves efficiency depending on the size of your data and your memory constraints. Let me know if you'd like further optimization in any particular direction!