Use re.findall to get valid letters

import re


def is_isogram(phrase):
    scrubbed = "".join(re.findall("[a-zA-Z]", phrase)).lower()
    return len(set(scrubbed)) == len(scrubbed)

For this approach, regular expression pattern, also known as a regex, is used to filter the input phrase string.

• In the pattern of [a-zA-Z] the brackets are used to define a character set that looks for characters which are a through z or A through Z. This essentially matches any characters which are in the English alphabet. The pattern is passed to the findall() method to return a list of matched characters.
• The result of findall() is passed as an argument to the join() method, which is called on an empty string. This makes a string out of the list of matched characters.
• The output of join() is then chained as the input for lower(). All of the letters are lowercased so that letters of different cases will become the same letter for comparison purposes, since A and a are considered to be the same letter. When the filtering and lowercasing is done, the scrubbed variable will be a string having all alphabetic letters lowercased.
• A set is constructed from the scrubbed string and its len is compared with the len of the the scrubbed string. Since a set holds only unique values, the phrase will be an isogram if its number of unique letters is the same as its total number of letters. The function returns whether the number of unique letters equals the total number of letters.
• For Alpha it would return False, because a is considered to repeat A, so the number of unique letters in Alpha is 4, and the total number of letters in Alpha is 5.
• For Bravo it would return True, since the number of unique letters in Bravo is 5, and the total number of letters in Bravo is 5.