Alphabet position in python

Question

Newbie here...Trying to write a function that takes a string and replaces all the characters with their respective dictionary values. Here is what I have:

def alphabet_position(text):

    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'}
    text = text.lower()
    for i in text:
        if i in dict:
            new_text = text.replace(i, dict[i])
    print (new_text)

But when I run:

alphabet_position("The sunset sets at twelve o' clock.")

I get:

the sunset sets at twelve o' cloc11.

meaning it only changes the last character in the string. Any ideas? Any input is greatly appreciated.

`new_text` keeps being replaced by the latest result. – mckuok Dec 07 '18 at 01:01 — mckuok, Dec 07 '18 at 01:01

Julien · Accepted Answer · 2018-12-07T01:14:38.050

Following your logic you need to create a new_text string and then iteratively replace its letters. With your code, you are only replacing one letter at a time, then start from scratch with your original string:

def alphabet_position(text):

    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'}
    new_text = text.lower()
    for i in new_text:
        if i in dict:
            new_text = new_text.replace(i, dict[i])
    print (new_text)

And as suggested by Kevin, you can optimize a bit using set. (adding his comment here since he deleted it: for i in set(new_text):) Note that this might be beneficial only for large inputs though...

I didn't see this answer when I posted and edited. Just deleted mine and voted for yours — Cory L, Dec 07 '18 at 01:13

score 2 · Answer 2 · answered Dec 07 '18 at 01:34

As your question is generally asking about "Alphabet position in python", I thought I could complement the already accepted answer with a different approach. You can take advantage of Python's string lib, char to int conversion and list comprehension to do the following:

import string


def alphabet_position(text):
    alphabet = string.ascii_lowercase
    return ''.join([str(ord(char)-96) if char in alphabet else char for char in text])

Paul Rooney · Answer 3 · 2018-12-07T11:38:12.200

Your approach is not very efficient. You are recreating the string for every character.

There are 5 e characters in your string. This means replace is called 5 times, even though it only actually needs to do anything the first time.

There is another approach that might be more efficient. We cant use str.translate unfortunately, as it's remit is one to one replacements.

We just iterate the input and produce a new string character by character.

def alphabet_position2(text):

    d = {L: str(i) for i, L in enumerate('abcdefghijklmnopqrstuvwxyz', 1)}

    result = ''
    for t in text.lower():
        result += d.get(t, t)

    return result

score 0 · Answer 4 · answered Nov 12 '19 at 13:21

This is a pretty simple approach with list comprehension.

Generate k:v in this format from string module, 1:b instead of b:1

import string

def alphabet_position(text):
        alphabeths = {v: k for k, v in enumerate(string.ascii_lowercase, start=1)}
        return " ".join(str(alphabeths.get(char)) for char in text.lower() if char in alphabeths.keys())

Alphabet position in python

4 Answers4