How can I add a hyphen between all even numbers of a value?

Question

Good morning, could someone help me with a code please, I need to insert a hyphen (-) between two even numbers. For example, if you receive the number 02368859 as input, the output of the program should be: 0-236-8-859 I did so, but you are in error.

num = input("Digite um número:")

def separador(num):
    num = list(num)
    fim = 1
    i = 0
    saida=''
    while i <len(num):
        if int(num[i])%2==0:
            saida+=num[i]
            if int(num[fim])%2==0:
                 saida+='-'
            i+=1
            fim+=1
        else:
            saida+=num[i]
            i+=1
            fim+=1        
    print(saida)

separador(num)

Answer 1

Regular Expressions

You can do this with regular expressions:

([02468]{2,})

Captures any sequence formed by the numbers 0, 2, 4, 6 or 8 with two or more of size. After capturing the sequences of even numbers just replace by itself by adding the hyphen:

'-'.join(grupo_capturado)  # '-'.join('28') = '2-8'

So the code stays:

import re

def adiciona_separador(matches):
    return '-'.join(matches.group(0))

resultado = re.sub(r'([02468]{2,})', adiciona_separador, '02368859')

print(resultado)  # 0-236-8-859

If you want to do a function for that:

def separador(numero, separador='-'):
    return re.sub(
        r'([02468]{2,})', 
        lambda matches: separador.join(matches.group(0)), 
        numero
    )

print( separador('02368859') )  # 0-236-8-859

^{See working on Repl.it | Ideone}

---

Forma Iterativa

Iteratively, as you tried to do, you can go directly through the string, do not need to convert it to a list. In Python strings are iterable as well. You can iterate over the characters in this string adding them into a string and whenever you find an even number you check whether the last value of this string is also even, if yes, you add a hyphen before adding the current number.

def separador(entrada):
    resultado = ''

    for caractere in entrada:
        if caractere in '02468':
            if len(resultado) > 0 and resultado[-1] in '02468':
                resultado += '-'
        resultado += caractere

    return resultado

print(separador('02368859'))  # 0-236-8-859

^{See working on Repl.it | Ideone}

Answer 2

You can use a purely regex approach. The "cat jump" is

• Use Lookahead to not lose overlay patterns.
• Use backreference to reinsert found digits

Example:

import re

resultado = re.sub(r'([02468])((?=([02468])))', r'\1-\2', '02368859')
print(resultado)

Read more on:

• https://docs.python.org/3/howto/regex.html#Lookahead-assertions
• https://docs.python.org/3/howto/regex.html#grouping

Answer 3

I found another approach in pure Python, quite similar to @Andersoncalos iteration code.

In it, I repeat each item from 0 to penultimate, adding this number to the output. And I check that both item "a" (current iteration index) and item "b" (current index + 1) are pairs. If so, I also add the "-" between them.

This would be function:

def separador2(num):
    saida = ''
    pares = '02468'
    for indice in range(len(num)-1):
        saida += num[indice] # Já adiciono o algarismo
        if num[indice] in pares and num[indice+1] in pares: # Se o "indice+1" for par também...
            saida += '-'
    saida += num[-1] #Adiciono o último caractere que eu omiti
    return saida

assert separador2('02368859') == '0-236-8-859'

Answer 4

I did it the way @Breno mentioned, but with small modifications to adapt to the proposed exercise.

num = input("Informe os valores:")
def separador(num):
saida = ''
pares = '02468'

for x in range(len(num)-1):
    saida += num[x]
    if num[x] in pares and num[x+1] in pares:
        saida += '-'
saida += num[-1]
print(saida)


separador(num)