Better way?

Question

Hello to everyone! What do you think about this code? Is there a better way to do this?

dict = {"a":1, "b":2, "c":3, "d":4, "e":5,
"f":6, "g":7, "h":8, "i":9, "j":10,
"k":11, "l":12, "m":13, "n":14, "o":15,
"p":16, "q":17, "r":18, "s":19, "t":20,
"u":21, "v":22, "w":23, "x":24, "y":25, "z":26}

palabra = input("Palabra: ").lower()
palabraCifrada =""

for letra in palabra:
if letra in dict:
palabraCifrada += str(dict[letra])
else:
palabraCifrada += letra

print(palabraCifrada)

Answer 1

You can use comprehension to build the dictionary so you don't need to write it manually (helps reduce typo too). And use string::replace method to replace each character that exists in the dictionary by the dictionary item value, respectively.

And as much as possible, avoid using class name as variable (`d` rather than `dict` for example).

Answer 2

# Lucio Fontana what about:

min = ord('a')-1

print(''.join(str(ord(chr)-min) if 'a' <= chr <= 'z' else chr for chr in palabra))

# avoid dict use/declaration
# use only palabra lowered user input

"""
however, this produce exactly same output as yours... but this is an one way encoding: you cannot decode output, unless you pass letters with 2 digits padded values and prefix original digits...
"""