Select data according to a specific time

Question

I am doing a program in which, given the day of the week, the hours and the current minutes, he responds with all the "Doctors" available at that time.

Each doctor has a period of care. For example:

• Carlos attends on Mondays from 10:00 to 12:00, and on Thursdays from 15:00 to 16:30.
• Juliana attends on Mondays from 10:00 to 11:30, and on Fridays from 15:00 to 16:00.

If someone runs the program at 11:00 on Monday, the program must print that Carlos and Juliana are available. If it runs on Wednesday at 15:00, the program should warn that no one is available. And finally, in case it runs on Thursday at 15:00, must return that Carlos is available.

I am doing this using Python 3. My logic is a sequence of if and else with their schedules, but I’m pretty sure it’s a pretty dumb way to solve that problem.

   from datetime import datetime
   from datetime import date
   
 async def responder(ctx):
    hoje = date.today()
    now = datetime.now()
    
    diaDaSemana = date.today()
    hora = now.hour
    minuto = now.minute
    
    if(diaDaSemana == 0):
        resposta = segundaFeira(hora, minuto)
    
    
    response = random.choice(brooklyn_99_quotes)
    await ctx.send(response)

def segundaFeira(hora, minuto):
    str = "Disponiveis: "
    if hora => 10 and hora <= 12:
        str += "Carlos"
    return str

Surely there is a simpler way to do this. I would like to know what this way is, I can use any python library.

Answer 1

An alternative is to store the available schedules and the respective doctors in a dictionary. A suggestion is to have this structure:

from datetime import datetime, time

# dias da semana: 0 - segunda, 1 - terça, etc
horarios = {
    'Carlos': {
        0: (time(10, 0), time(12, 0)), # segunda, das 10h às 12h
        3: (time(15, 0), time(16, 30)) # quinta, das 15h às 16:30
    },
    'Juliana': {
        0: (time(10, 0), time(11, 30)), # segunda, das 10h às 11:30
        5: (time(15, 0), time(16, 0)) # sexta, das 15h às 16h
    }
}

That is, for each name, I have another dictionary containing the days of the week (using the values 0 for Monday, 1 for Tuesday, and so on). I did so to be compatible with the values returned by datetime.weekday().

And for each day of the week, I create a tuple with the initial and final hours. With this, just go through the structure and see if the time fits in one of them:

from datetime import datetime, time

# dias da semana: 0 - segunda, 1 - terça, etc
horarios = {
    'Carlos': {
        0: (time(10, 0), time(12, 0)), # segunda, das 10h às 12h
        3: (time(15, 0), time(16, 30)) # quinta, das 15h às 16:30
    },
    'Juliana': {
        0: (time(10, 0), time(11, 30)), # segunda, das 10h às 11:30
        5: (time(15, 0), time(16, 0)) # sexta, das 15h às 16h
    }
}

# data e hora atual
hoje = datetime.now()
dia_da_semana = hoje.weekday() # obter o dia da semana
horario = hoje.time() # pegar somente o horário

# imprime a data atual
print(f'Hoje é {hoje:%A, %d/%m/%Y %H:%M:%S}')
encontrado = False
for nome, disponibilidade in horarios.items(): # para cada médico
    if dia_da_semana in disponibilidade: # se atende no dia da semana
        inicio, fim = disponibilidade[dia_da_semana]
        if inicio <= horario <= fim: # verifica se horário está entre o início e fim
            print(f'{nome} está disponível das {inicio:%H:%M} às {fim:%H:%M}')
            encontrado = True

if not encontrado:
    print('Não há médicos disponíveis')

---

The detail is that when printing the day of the week, by default the name is shown in English. To print in Portuguese, you need to use the module locale:

import locale
# setar locale para português
locale.setlocale(locale.LC_ALL, 'pt_BR.utf8')

# restante do programa é igual

Recalling that in this case the locale must be installed in the system, as explained in this reply.

And I used f-strings to print the date and times. But if you are using a Python version earlier than 3.6 an alternative is to use format along with strftime. Just change the print's for:

print('Hoje é {}'.format(hoje.strftime('%A, %d/%m/%Y %H:%M:%S')))

print('{} está disponível das {} às {}'.format(nome, inicio.strftime('%H:%M'), fim.strftime('%H:%M')))

---

I noticed that you are assembling a string with the available names. In this case, a solution is to go storing the names in a list, and at the end you join them in a string:

disponiveis = []
for nome, disponibilidade in horarios.items():
    if dia_da_semana in disponibilidade:
        inicio, fim = disponibilidade[dia_da_semana]
        if inicio <= horario <= fim:
            disponiveis.append(nome)

if disponiveis: # se encontrou alguém
    resposta = f'Disponíveis: {", ".join(disponiveis)}'
else: # não encontrou ninguém disponível
    resposta = 'Não há médicos disponíveis'

# usar a resposta aqui...

I created the list disponiveis containing the names of the doctors found. Then, if you have any, I put together the names using join (in the above example I used ", ", so names are separated by comma and space, but you can switch to whatever you want). The result is a string containing the names of all doctors found, which will be in the variable resposta.

---

More than one time on the same day

If you have more than one schedule in the same day, then it changes a little. In this case you need to have a list of tuples (each containing the start and end time). Something like this:

from datetime import datetime, time

horarios = {
    'Carlos': {
        0: [(time(10, 0), time(12, 0)), (time(15, 0), time(16, 0))], # segunda, das 10h às 12h e das 15h às 16h 
        3: [(time(15, 0), time(16, 30))] # quinta, das 15h às 16:30
    },
    'Juliana': {
        0: [(time(10, 0), time(11, 30)), (time(15, 30), time(17, 0))], # segunda, das 10h às 11:30 e das 15:30 às 17h
        5: [(time(15, 0), time(16, 0))] # sexta, das 15h às 16h
    }
}

hoje = datetime.now()
dia_da_semana = hoje.weekday() # obter o dia da semana
horario = hoje.time() # pegar somente o horário

disponiveis = []
for nome, disponibilidade in horarios.items():
    if dia_da_semana in disponibilidade:
        # procurar na lista de horários
        for inicio, fim in disponibilidade[dia_da_semana]:
            if inicio <= horario <= fim:
                disponiveis.append(nome)
                break # encontrei horário nesse dia, posso interromper o loop