Scrub with a series of replace calls

Isogram
Isogram in Python 
def is_isogram(phrase):
    scrubbed = phrase.replace('-', '').replace(' ', '').lower()
    return len(scrubbed) == len(set(scrubbed))

For this approach, replace() is called a couple times to scrub the input phrase string. Thw two replace() calls are chained, so the output of the first replace() is the input for the next replace(). The output of the last replace() is 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 replacing and lowercasing is done, the scrubbed variable will be a string having no hyphens or spaces, and with 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.
4th Sep 2024 · Found it useful?

Other Approaches to Isogram in Python

Other ways our community solved this exercise
def is_isogram(phrase):
    scrubbed = [ltr.lower() for ltr in phrase if ltr.isalpha()]
    return len(set(scrubbed)) == len(scrubbed)
Scrub with a list comprehension

Use a list comprehension 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.