The challenge

Given a string, replace every letter with its position in the alphabet.

If anything in the text isn’t a letter, ignore it and don’t return it.

"a" = 1, "b" = 2, etc.

Example

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

Should return "20 8 5 19 21 14 19 5 20 19 5 20 19 1 20 20 23 5 12 22 5 15 3 12 15 3 11" (as a string)

The solution in Python code

Option 1:

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def alphabet_position(text):
    return ' '.join(str(ord(c) - 96) for c in text.lower() if c.isalpha())

Option 2:

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def alphabet_position(text):
  al = 'abcdefghijklmnopqrstuvwxyz'
  return " ".join(filter(lambda a: a != '0', [str(al.find(c) + 1) for c in text.lower()]))

Option 3:

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alphabet = 'abcdefghijklmnopqrstuvwxyz'

def alphabet_position(text):
    if type(text) == str:
        text = text.lower()
        result = ''
        for letter in text:
            if letter.isalpha() == True:
                result = result + ' ' + str(alphabet.index(letter) + 1)
        return result.lstrip(' ')

Test cases to validate our solution

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from random import randint

test.assert_equals(alphabet_position("The sunset sets at twelve o' clock."), "20 8 5 19 21 14 19 5 20 19 5 20 19 1 20 20 23 5 12 22 5 15 3 12 15 3 11")
test.assert_equals(alphabet_position("The narwhal bacons at midnight."), "20 8 5 14 1 18 23 8 1 12 2 1 3 15 14 19 1 20 13 9 4 14 9 7 8 20")

number_test = ""
for item in range(10):
    number_test += str(randint(1, 9))
test.assert_equals(alphabet_position(number_test), "")