Convert alphabet letters to number in Python

Question

How can the following be finished?

characters = [\'a\'\'b\'\'c\'\'d\'\'e\'\'f\'\'g\'\'h\'\'i\'\'j\'\'k\'\'l\'\'m\'\'n\'\'o\'\'p\'\'q\'\'r\'\'t\'\'u\'\'v\'\'w         


        

Answer 1

Here's something I use to convert excel column letters to numbers (so a limit of 3 letters but it's pretty easy to extend this out if you need more). Probably not the best way but it works for what I need it for.

def letter_to_number(letters):
    letters = letters.lower()
    dictionary = {'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}
    strlen = len(letters)
    if strlen == 1:
        number = dictionary[letters]
    elif strlen == 2:
        first_letter = letters[0]
        first_number = dictionary[first_letter]
        second_letter = letters[1]
        second_number = dictionary[second_letter]
        number = (first_number * 26) + second_number
    elif strlen == 3:
        first_letter = letters[0]
        first_number = dictionary[first_letter]
        second_letter = letters[1]
        second_number = dictionary[second_letter]
        third_letter = letters[2]
        third_number = dictionary[third_letter]
        number = (first_number * 26 * 26) + (second_number * 26) + third_number
    return number

Answer 2

Use this function. It converts a string of alphabet to its equivalent digit value:

def convAlph2Num(sent):
    alphArray = list(string.ascii_lowercase)
    alphSet = set(alphArray)
    sentArray = list(sent.lower())
    x = []
    for u in sentArray:
        if u in alphSet:
            u = alphArray.index(u) + 1
            x.append(u)
    print(x)
    return

Answer 3

If you are going to use this conversion a lot, consider calculating once and putting the results in a dictionary:

>>> import string
>>> di=dict(zip(string.letters,[ord(c)%32 for c in string.letters]))
>>> di['c'] 
3

The advantage is dictionary lookups are very fast vs iterating over a list on every call.

>>> for c in sorted(di.keys()):
>>>    print "{0}:{1}  ".format(c, di[c])
# what you would expect....

Answer 4

import string
# Amin
my_name = str(input("Enter a your name: "))
numbers      = []
characters   = []
output       = []
for x, y in zip(range(1, 27), string.ascii_lowercase):
    numbers.append(x)
    characters.append(y)

print(numbers)
print(characters)
print("----------------------------------------------------------------------")

input = my_name
input = input.lower()

for character in input:
    number = ord(character) - 96
    output.append(number)
print(output)
print("----------------------------------------------------------------------")

sum = 0
lent_out = len(output)
for i in range(0,lent_out):
    sum = sum + output[i]

print("resulat sum is : ")
print("-----------------")

print(sum)





resualt is :
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26]
['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v', 'w', 'x', 'y', 'z']
----------------------------------------------------------------------
[1, 13, 9, 14]
----------------------------------------------------------------------
resulat sum is : 
-----------------
37