I want to change a string in 3 positions

If I understand you, you want to do something like Cesar's cipher. Given a string, you need to walk with each character plus 3 positions.

You can use a basic alphabet, in python it is already implemented by default in string

from string import ascii_lowercase as alfabeto

so you can have a base to rotate. Beauty, we have a base of strings. So let's get a string to put in the game:

string = 'stack'

The string has a method called .index that will give you the position of each character. For example:

alfabeto.index('a')
# 0

so you can add 3 more and get a new position. In the same case:

alfabeto[alfabeto.index('a') + 3]
# 'd'

In this case, you can iterate through the string you want to rotate

for letra in 'stack':
    print(alfabeto[alfabeto.index(letra) + 3)
# 'v'
# 'w'
# 'd'
# 'f'
# 'n'

This way you would have a whole string rotated. However, there are some problems, such as if you try to rotate 'z'

for letra in 'zzz':
    print(alfabeto[alfabeto.index(letra) + 3])
# IndexError: string index out of range

Because your string does not have a position greater than 'z'. So you could make a module using the size of your alphabet.

for letra in 'zzz':
    print(alfabeto[(alfabeto.index(letra) + 3)% len(alfabeto)])
#'c'
#'c'
#'c'

What would you always run the whole alphabet by adding three more and starting with the same case it pops the size of your list.

''.join([alfabeto[(alfabeto.index(letra) + 3)% len(alfabeto)] for letra in 'stackzz'])
# 'vwdfncc'

You have to convert to number to make the account (with ord() ) and then convert to character again (with chr() ), like this:

def DeslocaASCII(texto):
    novoTexto = ''
    for letra in texto:
        numero = ord(letra)
        novoTexto += chr(numero + (-23 if numero > 87 else 3))
    return novoTexto

print(DeslocaASCII('ABCDEFGHIJKLMNOPQRSTUVWXYZ'))

See running on ideone . And no Coding Ground . Also put it in GitHub for future reference .

I'm considering that only text with upper-case characters will be sent. Validation of input text should be done elsewhere to maintain sole responsibility.

I have not tried to solve anything beyond what is in the question so as not to speculate on where it will be used.

def rot(s):
   return ''.join([chr(ord('A')+(ord(c)-ord('A')+3)%26) for c in s])

• Given a letter c , p=ord(c)-ord('A') , gives its position within the letters. p ∈ {0..25};
• (p+3)%26 sum 3 circularly within segment [0..25] ;
• chr(p + ord('A')) gives the letter corresponding to the position p ;
• ord('A')+(ord(c)-ord('A')+3)%26 sum 3 circulamente within the section ['A' .. 'Z']
• [ soma3(c) for c in s] computes the list of rounded letters present in s
• ''.join(lista) rebuilds the already-rotated string.

Lovers of unreadable answers can still replace ord ('A') by getting:

def r(s): return ''.join([chr(65+(ord(c)-10)%26) for c in s])