I want to change a string in 3 positions

Question

I would like to walk 3 positions with a letter of a string, for example, make the letter A turn D, I tried the second command:

texto[c] = texto[c] + 3

But it still doesn’t work, what would be the right way to do?

Answer 1

If I get it, you want to do something similar to Cesar’s cipher. Given a string, you need to walk with each character plus 3 positions.

You can use a base 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 string base. So let’s take a string to put in the game:

string = 'stack'

String has a method called .index which 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 on 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 fully rotated string. However, there are some problems, 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 larger than 'z'. Then 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 make you always run the whole alphabet adding up three more and starting with the same case it bursts the size of your list.

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

Answer 2

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

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'))

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

I am considering that only a text with uppercase 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 not to speculate where it will be used.

Answer 3

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 inside the letters. p {0.. 25};
• (p+3)%26 sum 3 circularly within the 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 circularly within the quarter ['A'.. 'Z']
• [ soma3(c) for c in s] calculates the list of worked letters present in s
• ''.join(lista) rebuilt the string already round.

Lovers of unreadable answers may also replace Ord('A') getting:

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