[Django]-Get minimum value field name using aggregation in django

If you pass the Min as positional argument, then the field’s name is price__min. Otherwise, if you pass it as keyword argument, i.e. aggregate(my_min=Min('price')), then it will be available with the same name as the argument, in this case my_min. Docs

For Min and Max you may order your values (QuerySet result) and grab the first() and last():

chocolate_list = Chocolate.objects.values_list('name', 'price')
min = chocolate_list.order_by('price').first()
max = chocolate_list.order_by('price').last()

PS: Remove the filter() if you are not assigning nothing. With values_list() you are implicitly instancing the QuerySet class

Usage of what @Maciej Gol said:

from django.db.models import Min    

lowest_price = Chocolate.objects.values('price').aggregate(Min('price'))['price__min']

Expanding on Jcc’s answer and Luciano’s comment, the solution for the exact OP’s question would be the following.

Code

cheapest_choclate = Choclate.objects.values_list('name', 'price').order_by('price')[0]
lowest_price = cheapest_choclate['price']
cheapest_choclate_name = cheapest_choclate['name']

Explanation

Using Choclate.objects, you obtain all objects of class Choclate from the DB.

Using .values_list('name', 'price'), you create a QuerySet with a list of tuples where each tuple contains the ‘name’ and ‘price’ of each object – e.g. [('Sweet', 79),('Bitter', 49), ('Spicy', 69)].

Using .order_by('price'), you order the tuples in the list using the price – e.g. [('Bitter', 49), ('Spicy', 69), ('Sweet', 79)].

Using [0], you select the first tuple from the list. This tuple contains the name and price of the cheapest choclate. You could also use .first() as suggested by Jcc. However, I like the [0] approach more as it can easily be changed to something else if needed – e.g. "second cheapest" choclate could be obtained using [1].

Lastly, you can obtain the price and name of the cheapest choclate from the first tuple using cheapest_choclate['price'] and cheapest_choclate['name'].