[Django]-Django Aggregation: Sum return value only?

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.

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.

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)

(It’s only useful if you used decimal field)
Store a key value in a new variable and convert it to the float if you’re using a decimal. I think it’s the easiest way

total_paid = Payment.objects.all.aggregate(Sum('amount'))
# output: {'amount__sum':Decimal('0000.000000')
tp=total_paid['amount__sum']
new_total_paid=float(tp)

100% you can solve your demand by this technique

from django.db.models import Sum

sum_count = Payment.objects.aggregate(Sum('amount'))
totalAmonut = sum_count['Qty__sum']