How to verify and print repeating values in an algorithm vector

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Create an algorithm that given a sequence of n real numbers, determine the numbers that make up the sequence and the number of times each of them occurs in it.

Example: n = 8

Sequel:

-1.7,  
3.0,  
0.0,  
1.5,  
0.0,  
-1.7,  
2.3,  
-1,7

Exit:

-1.7 ocorre 3 vezes

 3.0 ocorre 1 vez

 0.0 ocorre 2 vezes

 1.5 ocorre 1 vez

 2.3 ocorre 1 vez

I did it in the Visualg Code:

ALGORITMO "QUESTAO 9"
VAR
C,X,SEQ,I: INTEIRO
VSR: VETOR[1..100] DE REAL
SEQUENCIA: VETOR[1..100] de REAL

INICIO
ESCREVA("QUANTOS NÚMEROS REAIS COMPÕEM A SEQUÊNCIA? ")
LEIA(SEQ)
ESCREVAL("INFORME A SEQUÊNCIA: ")
PARA I DE 1 ATE SEQ FACA
LEIA(SEQUENCIA[I])
SE (SEQUENCIA[I])<>(VSR[I]) então
VSR[I]<-SEQUENCIA[I]
fimse
fimpara
para X de 1 ate SEQ faca
C <- 0
para I de 1 ate SEQ faca
se (VSR[X] = SEQUENCIA[I]) então
C <- C + 1
fimse
fimpara
escreval(VSR[X], " OCORRE ", C, " VEZES")
fimpara
escreval(" ")
escreval(VSR[X], " OCORRE ",SEQUENCIA[SEQ], " VEZES")


fimalgoritmo
algorithm visualg

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  • 1

    This answers your question? Get how many times each element of a list repeats

  • Unless a typo has occurred the value -1.7 occurs twice, not three, in your list of 9 elements.

  • The method Counter of lib collections fix it. See here

  • @Overpeas, it’s hard to see the code on comments. Update your post, please.

1 answer

0


I think that’s what you seek.

Update

N = int(input("Quantos números? "))

lista = []
for i in range(1, N+1):
    lista.append(float(input(f"{i} - Entre com o número: ")))

NOTE: Typing errors are not being handled

from collections import Counter

c = Counter(lista)

Example of Results

print(f"Lista de numeros: {list(c.keys())}")

Lista de numeros: [-1.7, 3.0, 0.0, 1.5, 2.3]

for numero, repeticoes in c.items():
    print(f" O numero {numero} repete {repeticoes} vezes")

 O numero -1.7 repete 3 vezes
 O numero 3.0 repete 1 vezes
 O numero 0.0 repete 2 vezes
 O numero 1.5 repete 1 vezes
 O numero 2.3 repete 1 vezes

I hope it helps

  • That’s right @Paulo Marques, thank you very much.

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