Hola, puedes utilizar un for y guardar la salida en un query aunque por optimización los queries de django solo se ejecutan en el momento de usarlos entonces podrias crear un templetag para esto.
def cliente(request):
...
...
clientes = Clientes.objects.order_by('Nombre')
queries = {}
for a in range(97, 123):
queries[a] = clientes.filter(Nombre__starswith=a)
...
...
print queries['f']
print queries['z']
>
> def clientes(request):
> clientes = Clientes.objects.order_by('Nombre')
> clientes2 = Clientes.objects.order_by('Nombre').filter(medio='1')
> clientes3 = Clientes.objects.order_by('Nombre').filter(medio='2')
> clientes4 = Clientes.objects.order_by('Nombre').filter(medio='3')
> c = Clientes.objects.filter( Nombre__startswith='a').order_by('Nombre')
> template = 'products/Clientes.html'
> return render(request, template, locals())
>
> tengo el siguiente problema quiero filtrar por la letra del abecedario y lo
> realizo con
>
> c = Clientes.objects.filter( Nombre__startswith='a').order_by('Nombre')
>
> pero tengo uno para cada letra nose si me puedan ayudar con alguna funcion
> que me facilite obtener esto
>
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