How do I show the wrong input numbers?

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1

In this code I would like that if the student enters some wrong note, the program responds which grades are wrong. But based on what I did only appears the first note.

I also wish that when typing the correct notes appear whether it was approved or not.

n1 = int(input('digite nota 1: '))
n2 = int(input('digite nota 2: '))
media = (n1 + n2)/2
if n1 > 10:
  print('{} não é uma nota válida!'.format(n1))
elif n2 > 10:
  print('{} não é uma nota válida!'.format(n2))
elif media >= 7 and media<=10:
  print('Você foi aprovado com média {}'.format(media))
elif media < 7 and media<=10:
  print('Você foi reprovado com média {}'.format(media))
python algorithm

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2 answers

2

Are you abusing the elif. It is not because it has several conditions that everything should be grouped together. It has to do this according to the need.

The validation of each note should be independent, it makes no sense to put a elif there, except to complicate the code, it could in a different way but it complicates the code a lot and it would be less readable. The most correct when an invalid data is entered is to close. If it were in a function it would be enough to use a return, if it were in a loop, could use a continue or even a break in some situations. As in script pure must come out with exit().

The average should only be calculated when it has valid data and continued execution.

N]ao it is necessary to check if the average is up to 10 because if the data are valid it has been calculated to the correct limit.

And finally you don’t need to check anything on the disapproval because it’s the opposite of approval, just one else simple.

import sys

n1 = int(input('digite nota 1: '))
n2 = int(input('digite nota 2: '))
if n1 > 10:
    print('{} não é uma nota válida!'.format(n1))
    sys.exit(0)
if n2 > 10:
    print('{} não é uma nota válida!'.format(n2))
    sys.exit(0)
media = (n1 + n2) / 2
if media >= 7:
    print('Você foi aprovado com média {}'.format(media))
else:
    print('Você foi reprovado com média {}'.format(media))

Behold working in the ideone. And in the repl it.. Also put on the Github for future reference.

Can you do it the way you were doing without forcing yourself out? Yes, but logic would be too confusing for an experienced person to imagine for a starting person. And I would risk learning in a confusing way for the rest of my life. Otherwise it would be less efficient, although it is not something relevant to this case could be one day and already accustomed to do more correct.

-1

You can build a function able to read the notes, treat them according to the rules set, display any errors and only return if the typed note is valid, see only:

def obter_nota_aluno(msg):
    while True:
        try:
            # Lê a nota
            n = input(msg)

            # Converte e verifica se a nota é um inteiro válido
            n = int(n) 

            # Verifica se a nota está dentro da faixa válida
            if (n < 0) or (n > 10):
                raise ValueError()

            # Retorna nota
            return n
        except ValueError:
            # Exibe o erro
            print(f'{n} não é uma nota válida, tente novamente.')

n1 = obter_nota_aluno('digite nota 1: ')
n2 = obter_nota_aluno('digite nota 2: ')

media = (n1 + n2) / 2

if media >= 7:
    print(f'Você foi APROVADO com média {media}')
else:
    print(f'Você foi REPROVADO com média {media}')

See working on Repl.it

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