I have a model:
class ItemPrice(models.Model):
price = models.DecimalField(max_digits=8, decimal_places=2)
# ...
I tried this to calculate the sum of price in this queryset:
items = ItemPrice.objects.all().annotate(Sum('price'))
What's wrong in this query? or is there any other way to calculate the Sum of price column?
I know this can be done by using for loop on queryset but i need an elegant solution.
Thanks!
You're probably looking for aggregate
from django.db.models import Sum
ItemPrice.objects.aggregate(Sum('price'))
# returns {'price__sum': 1000} for example
{'price__sum':1000} . Can get float/integer with yourdict['price__sum']Use .aggregate(Sum('column'))['column__sum'] reefer my example below
sum = Sale.objects.filter(type='Flour').aggregate(Sum('column'))['column__sum']
Annotate adds a field to results:
>> Order.objects.annotate(total_price=Sum('price'))
<QuerySet [<Order: L-555>, <Order: L-222>]>
>> orders.first().total_price
Decimal('340.00')
Aggregate returns a dict with asked result:
>> Order.objects.aggregate(total_price=Sum('price'))
{'total_price': Decimal('1260.00')}
key to store the sum value into.Using cProfile profiler, I find that in my development environment, it is more efficient (faster) to sum the values of a list than to aggregate using Sum().
eg:
sum_a = sum([item.column for item in queryset]) # Definitely takes more memory.
sum_b = queryset.aggregate(Sum('column')).get('column__sum') # Takes about 20% more time.
I tested this in different contexts and it seems like using aggregate takes always longer to produce the same result. Although I suspect there might be advantages memory-wise to use it instead of summing a list.
sum_a = sum(item.column for item in queryset). The only difference is the removed []s. This saves the memory space for calculating the entire list before sum() iterates over it.Previous answers are pretty well, also, you may get that total with a line of vanilla code...
items = ItemPrice.objects.all()
total_price = sum(items.values_list('price', flat=True))
You could also get the sum this way:
def total_sale(self):
total = Sale.objects.aggregate(TOTAL = Sum('amount'))['TOTAL']
return total
Replace the 'amount' with the column name from your model you want to calculate the sum of and replace 'Sale' with your model name.
You need to use aggregate() and Sum() to calculate the sum of price column as shown below. *The query with all() is equivalent to the query without all() as shown below:
from django.db.models import Sum
print(ItemPrice.objects.all().aggregate(Sum('price')))
print(ItemPrice.objects.aggregate(Sum('price')))
Then, these dictionaries below are outputted on console:
{'price__sum': Decimal('150.00')}
{'price__sum': Decimal('150.00')}
And, you can change the default key price__sum to priceSum for price column as shown below:
from django.db.models import Sum
# ↓ Here
print(ItemPrice.objects.all().aggregate(priceSum=Sum('price')))
print(ItemPrice.objects.aggregate(priceSum=Sum('price')))
# ↑ Here
Then, the default key is changed as shown below:
{'priceSum': Decimal('150.00')}
{'priceSum': Decimal('150.00')}
You can also make it simple
from django.db.models import Sum
product_total= Product.objects.aggregate(price=Sum('price'))['price'] or 0
which will give you just the amount. Eg: 4379 so the price is from your database column
