Is there a shorter way to replace all letters in an input?

I am making a thing that needs to replace all letters in the alphabet taken from a user's input but I think just doing it this way is way too long. Is there a better way to do this?

Im using python 3, b in the code is the input, and dot is the name of the list with stuff to replace

a1 = b.replace("a", dot[0])
  a2 = a1.replace("b", dot[1])
  a3 = a2.replace("c", dot[2])
  a4 = a3.replace("d", dot[3])
  a5 = a4.replace("e", dot[4])
  a6 = a5.replace("f", dot[5])
  a7 = a6.replace("g", dot[6])
  a8 = a7.replace("h", dot[7])
  a9 = a8.replace("i", dot[8])
#.... Continues

CodePudding user response：

With the replacement in a dictionary we can use str.translate as long as you only need to replace single characters.

dotmap = {'e':'3', 'o':'0', ...}
txt = 'Hello world'

replacements = txt.maketrans(dotmap)
ntxt = txt.translate(replacements)

Which will return H3ll0 w0rld.

CodePudding user response：

I'm guessing dot contains the replacement characters for each alphabet. You could do something like

import string

s = 'example'
dot = '12345'

for c, cnew in zip(string.ascii_lowercase, dot):
    s = s.replace(c, cnew)

print(s)  # prints '5x1mpl5'

Explanation: zip pairs together the lowercase ASCII characters and the replacement characters. Then we get each pair in turn as c and cnew. s becomes the new string and the loop is repeated until either dot or the lowercase alphabet is exhausted.

CodePudding user response：

An idea is to convert the character into an integer, add 1 to it and then cast the resultant integer to char. You can achieve this using the builtin methods ord and chr.

prev=b

for i in range(26):
   cur=prev.replace(chr(ord('a') i), dot[i])
   prev=cur