如何使用python将列表索引值向前移动'x'次？

Question

I'm trying to unscramble a code using pythonic methods. The way to crack the code is by selecting the letter two places ahead of itself.

For example if the code was

abc

Then the solution would be

cde

So I am trying to figure out how I add 2 to the index value of a letter (given it is in a list like the following)

alphabet = ["a","b","c"...."z"]

I am trying to write code similar to this

def scramble(sentence):
    alphabet = ["a","b","c"...]
    solution = []
    for i in sentence:
        newIndex = get index value of i, += 2
        newLetter = translate newIndex into the corresponding letter
        solution.append(newLetter)
    for i in solution:
        print(i, end="")

But I am not yet skilled enough in python to figure it out

Answer 1

Try:

>>> s = 'abc'
>>> ''.join(chr(ord(c)+2) for c in s)
'cde'

The above is not limited to standard ASCII: it works up through the unicode character set.

Limiting ourselves to 26 characters

>>> s = 'abcyz'
>>> ''.join(chr(ord('a')+(ord(c)-ord('a')+2) % 26) for c in s)
'cdeab'

Modifying the original code

If we want to modify the original only enough to get it working:

from string import ascii_lowercase as alphabet

def scramble(sentence):
    solution = []
    for i in sentence:
        newIndex = (alphabet.index(i) + 2) % 26
        newLetter = alphabet[newIndex]
        solution.append(newLetter)
    for i in solution:
        print(i, end="")

Example:

>>> scramble('abcz')
cdeb

Answer 2

One solution would be to use enumerate:

for idx, char in enumerate(sentence):
    # Do what you need to do here, idx is the index of the char, 
    # char is the char itself
    letter = sentence[idx + 2] # would get you the char 2 places ahead in the list

This way you could index in by adding to idx. Make sure you check for the end of the list though.

You could also wrap around the list by using modulus on the length of the list, so if you added 2 to the index 25 in a 26 element list, you would get 27 % 26 = 1, so the second element of the list again.

Answer 3

There are a couple of ways to do this.

First one, use a dictionary.

a={'a':'c','b':'e'....}
sol=[]
for i in sentence:
    sol.append(a[i])

Answer 4

alphabet = "abc"
ord_a = ord('a')
solution = ''.join(chr((ord(c)-ord_a+2)%26+ord_a) for c in alphabet)

Answer 5

Something like this would be useful because it would allow you to change your offset value easily. It'll also handle things other than characters

def caesar_cipher(sentence, offset):
   pt_alphabet = ['a','b'...'z']
   ct_alphabet = []
   index = len(pt_alphabet) + offset
   while len(ct_alphabet) < len(pt_alphabet):
      if index < (len(pt_alphabet) - 1):
         ct_alphabet.append(pt_alphabet[index])
         index += 1
      else:
         index -= len(pt_alphabet)
   out_word = ''
   for character in sentence:
      if character.lower() not in pt_alphabet:
         out_word += character
      if character.lower() in pt_alphabet:
         out_word += ct_alphabet[pt_alphabet.index(character.lower())]
   return out_word

if __name__ == '__main__':
   offset = 2
   sentence = "This is a test with caps and special characters!!"
   cipher_text = caesar_cipher(sentence, offset)
   print sentence
   print cipher_text

Usage like this:

$ python cipher.py
>> This is a test with caps and special characters!!
>> vjku ku c vguv ykvj ecru cnf urgekco ejctcevgtu!!