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Django: resumir recuentos en una consulta

Tengo una lista de elementos que me gustaría clasificar según el recuento de muchos elementos relacionados. Creo que agregar los recuentos de esos elementos de origen es la forma correcta, pero aún no he encontrado una solución.

 class Element(models.Model): Source1 = models.ManyToManyField(Source1) Source2 = models.ManyToManyField(Source2) Source3 = models.ManyToManyField(Source3) Ranked = (Element.objects.all().aggregate( Ranked=Sum(F('Source1') + F('Source2') + F('Source3'), output_field=IntegerField)['Ranked'] ))
about 4 years ago · Santiago Trujillo
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Una posible solución:

 agg_data = Element.objects.aggregate(total_source1=Count('Source1'), total_source2=Count('Source2'), total_source3=Count('Source3')) total_count = sum(agg_data.values()) # this is value which you need

Actualización: si desea obtener una lista de elementos:

 res = Element.objects.all().annotate(total_source1=Count('Source1'), total_source2=Count('Source2'), total_source3=Count('Source3')) .annotate(total_count=F('total_source1') + F('total_source2') + F('total_source3')).order_by('-total_count')
about 4 years ago · Santiago Trujillo Report
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