Django Aggreagtion: Sum return value only?

Question

I have a list of values paid and want to display the total paid. I have used Aggregation and Sum to calculate the values together. The problem is,I just want the total value printed out, but aggreagtion prints out: {'amount__sum': 480.0} (480.0 being the total value added.

In my View, I have:

    from django.db.models import Sum

    total_paid = Payment.objects.all.aggregate(Sum('amount'))

And to show the value on the page, I have a mako template with the following:

    <p><strong>Total Paid:</strong> ${total_paid}</p>

How would I get it to show 480.0 instead of {'amount__sum': 480.0}?

Answer 1

I don't believe there is a way to get only the value.

You could just do ${{ total_paid.amount__sum }} in your template. Or do total_paid = Payment.objects.all().aggregate(Sum('amount')).get('amount__sum', 0.00) in your view.

EDIT

As others have pointed out, .aggregate() will always return a dictionary with all of the keys from the aggregates present, so doing .get() on the result is not necessary. However, if the queryset is empty, each aggregate value would be None. So depending on your code, if you are expecting a float, you could do:

total_paid = Payment.objects.all().aggregate(Sum('amount'))['amount__sum'] or 0.00

Answer 2

The aggregate() method returns a dictionary. If you know you're only returning a single-entry dictionary you could use .values()[0].

In Python 2:

total_paid = Payment.objects.aggregate(Sum('amount')).values()[0]

In Python 3, (thanks @lmiguelvargasf) this will need to be:

total_paid = list(Payment.objects.aggregate(Sum('amount')).values())[0]

The end result is the same as @jproffitt's answer, but it avoids repeating the amount__sum part, so it's a little more generic.

Answer 3

In Python 3:

You can solve it by converting the dict_values to a list:

total_paid = list(Payment.objects.aggregate(Sum('amount')).values())[0] or 0 # the or 0 is required in case the query is an empty query set.

The previous code avoids using 'column_name__sum' as key, but in case you prefer the dictionary way:

total_paid = Payment.objects.aggregate(Sum('amount'))['amount__sum'] or 0

In terms of efficiency, I made a test with some data I have, and it seems that using the dictionary key is faster:

In [9]: %timeit total = Pledge.objects.filter(user=user, group__isnull=True).aggregate(Sum('amount'))['amount__sum'] or 0
3.13 ms ± 25.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In [10]: %timeit total = list(Pledge.objects.filter(user=user, group__isnull=True).aggregate(Sum('amount')).values())[0] or 0
3.22 ms ± 61.1 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

In terms of readability, I think that @mehmet's solution is the best one, and I have also test its efficiency:

In [18]: %timeit Pledge.objects.filter(user=user, group__isnull=True).aggregate(sum=Sum('amount'))['sum'] or 0
3.22 ms ± 124 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 4

Give it a name and then ask for it:

total_paid = Payment.objects.all.aggregate(sum=Sum('amount'))['sum']

Should be little more readable, and there is no need for conversion.