How to save multiple read values to be used later

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Problem:

"Read a sequence of 1000 integers. Read another integer N, and your program should print how many times the integer N appears in the previous 1000. The program stops when the first integer of the 1000 is equal to -1. For each 1000 integer package, the following sentence should be printed: N appeared K times"

aux = 0
while aux < 1000
      aux = aux + 1
      a = input()
n = input()

How I fit the command for and how I keep the values of a to compare with each other later? I want to solve it this way because I always doubt it.

python

2 answers

1


Keep the numbers in one list:

numeros = [] # lista que vai guardar os números
for i in range(1000):
    n = int(input())
    if i == 0 and n == -1: # se o primeiro é -1, sai do loop
        break
    numeros.append(n) # adicionar o número na lista

I used range(1000), which generates the sequence of numbers between 0 and 999 (i.e., the for will iterate 1000 times). And I’ve included the stop rule when the first number is -1 (break interrupts the loop). And if this occurs, the list of numbers will be empty.

I also used int to turn what was typed into number (will give error if not typed a number, it is not clear if it is requirement of the problem).

To count how many times n occurs on the list:

if numeros: # se a lista não está vazia
    n = int(input())
    cont = 0
    for i in numeros:
        if i == n:
            cont += 1
    print(f'{n} appeared {cont} times')

I understand that because it is an exercise, they probably want you to do the counting manually, as it is above. But in any case, it already exists ready in the language, through the method count:

if numeros: # se a lista não está vazia
    n = int(input())
    print(f'{n} appeared {numeros.count(n)} times')

Another option to quit the program (since the statement says the program should stop) is to use sys.exit:

import sys
numeros = []
for i in range(1000):
    n = int(input())
    if i == 0 and n == -1: # se o primeiro é -1, para o programa
        sys.exit(0)
    numeros.append(n)

# não preciso mais verificar se a lista é vazia
n = int(input())
# etc...

Like sys.exit exits the program, I no longer need to check if the list is empty after the for.

And of course it can also be done with while instead of for:

i = 0
numeros = []
while i < 1000:
    n = int(input())
    if i == 0 and n == -1: # se o primeiro é -1, para o programa
        break # ou sys.exit(0)
    numeros.append(n)
    i += 1

And if the idea is to repeat several times the reading of the 1000 numbers, put everything within one loop infinite:

import sys
while True:
    numeros = []
    for i in range(1000):
        n = int(input())
        if i == 0 and n == -1: # se o primeiro é -1, para o programa
            sys.exit(0)
        numeros.append(n)

    n = int(input())
    cont = 0
    for i in numeros:
        if i == n:
            cont += 1
    print(f'{n} appeared {cont} times')

The while True is repeated until the sys.exit be called.

  • 1

    Thanks for the explanation! My way of using the while command wouldn’t work, right? You need to use 'for'''.

  • @Maria You can also do it with while, I updated the answer by adding this option.

  • Thank you very much! I understood perfectly!

  • Just one observation: the program enters 1000 numbers several times, until, in one of these entries, the first one is -1.

  • @Maria Atualizei a resposta

0

You can use an array to store values.

numeros = [] #cria o array
numeros.append(i) #add i ao array

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