IsAnagram? - Quick Hack

Hello folks,

Just sharing a quick hack to find out if two words are anagrams of each other or not.

def isAnagram(word,word2):
    return sorted(word.lower().strip().replace(" ","")) == sorted(word2.lower().replace(" ",""))
print(isAnagram("Dormitory","Dirty room"))
print(isAnagram("School master", "The classroom"))

This function returns a boolean output based on whether they're anagrams or not.

Using sorted() to compare two words in an orderly manner to check their anagrammatic property & it just makes things a lot easier! :)

• Added the ".lower()" to convert the uppercase to lowercase (if there's an upper case letter) and check since different letter cases impact the result.
• Also, using ".replace(" ","")" to remove whitespaces.

Code on -- Cheers!

Thanks to the brilliant comments, made some edits to change the logic from set() to sorted() to rule out false positives

Comments

[Kelton]

I'm curious as to why you used sets rather than a sorted list.

You said you're using sets because they "use an orderly arrangement of alphabets", I thought python sets are unordered.

Also, a set cannot have duplicate elements, so wouldn't "apple" and "leap" return True despite not being anagrams?

[Shyam Salil]

That’s a brilliant catch! Thank you. I will change this post to reflect that :)

[Matt Hamilton]

Using a set operator though you will remove duplicate letters. Which will give you some false positives.

[Shyam Salil]

Yup, made the edits :) Thank you! Might have never noticed the loop if it weren't for these comments.