Order a dict in python when keys are list

Question

I need to order the following dict in asc order depends on the progreso value. How can I do it?

{
    10: [{"titulo": "Apropiación Digital"}, {"progreso": "50"}, {"estado": "En curso"}],
    13: [
        {"titulo": "Así se ve mi negocio"},
        {"progreso": "0"},
        {"estado": "En espera"},
    ],
    8: [{"titulo": "Bioseguridad"}, {"progreso": "50"}, {"estado": "En curso"}],
    15: [
        {"titulo": "Desarrollo de oportunidades de negocio"},
        {"progreso": "0"},
        {"estado": "En espera"},
    ],
    9: [{"titulo": "Formalización"}, {"progreso": "0"}, {"estado": "En espera"}],
    11: [{"titulo": "Hagamos cuentas"}, {"progreso": "50"}, {"estado": "En curso"}],
    7: [
        {"titulo": "Mujer emprendedora"},
        {"progreso": "100"},
        {"estado": "Finalizado"},
        {"puntaje": 100},
        {"fecha": datetime.datetime(2021, 3, 31, 15, 19, 20)},
        {"puntos": 170},
    ],
    12: [
        {"titulo": "Precio y competencia"},
        {"progreso": "0"},
        {"estado": "En espera"},
    ],
    14: [{"titulo": "Servicio al cliente"}, {"progreso": "0"}, {"estado": "En espera"}],
    16: [{"titulo": "Test"}, {"progreso": "0"}, {"estado": "En espera"}],
}

Answer 1

We know the location of the progreso value in the dictionary:

{10: [ .. , {'progresso':' x '}, ..], 13: [ .. ,x, .. ], ..}

so in this dictionary, you first have the name (dict for example), then the first integer, then the second element ( progress )

while True:
    tempDict = dict               # The dictionary will be equal to the temporary dictionary at the end of this loop if there are no more changes made to it.
    for i in dict[0:-2]:          # From first to second last element, to prevent it from looking beyond the last element
        if dict[i]['progreso'] < dict[i+1]['progreso']:
             dict[i],dict[i+1] = dict[i+1],dict[i] #swap places 'i' and 'i+1'
    if tempDict == dict: # Now that changes may have been made, we can test for it, and possibly exit the loop.
        break

something along those lines.