Scrub with a list comprehension

Isogram
Isogram in Python 
def is_isogram(phrase):
    scrubbed = [ltr.lower() for ltr in phrase if ltr.isalpha()]
    return len(set(scrubbed)) == len(scrubbed)

For this approach, a list comprehension is used to iterate the letters in the input phrase string.

  • In the code example, ltr is the name given to each letter iterated in the for loop.
  • The result of each iteration is ltr.lower(), which is the lowercased letter being iterated. 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.
  • The iterable part of the list comprehension is the input phrase.
  • The letters are filtered by the use of the optional conditional logic: if ltr.isalpha(). isalpha() returns True if the letter being iterated is alphabetic.

When the list comprehension is done, the scrubbed variable will be a list holding only lowercased alphabetic characters.

  • A set is constructed from the scrubbed list and its len is compared with the len of the the scrubbed list. 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.
4th Sep 2024 · Found it useful?

Other Approaches to Isogram in Python

Other ways our community solved this exercise
def is_isogram(phrase):
    scrubbed = phrase.replace('-', '').replace(' ', '').lower()
    return len(scrubbed) == len(set(scrubbed))
Scrub with a series of replace calls

Use a series of replace calls and set to return the answer. 

import re


def is_isogram(phrase):
    scrubbed = re.compile('[^a-zA-Z]').sub('', phrase).lower()
    return len(set(scrubbed)) == len(scrubbed)
Scrub with a regex

Use a re.sub and set to return the answer. 

import re


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

Use a re.findall and set to return the answer. 

for ltr in phrase:
    letter = ord(ltr)
    if letter >= A_LCASE and letter <= Z_LCASE:
        if letter_flags & (1 << (letter - A_LCASE)) != 0:
            return False
        else:
            letter_flags |= 1 << (letter - A_LCASE)
Bit field using a for loop

Use a bit field with a for loop to keep track of used letters.