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pythonfunctionif-statementconditional-statementsanagram
frustration
frustration

Reputation: 63

Creating an anagram checker

So I have been able to create the following program which compares two strings to see if they are anagrams of each other. 

def anagrams( string1, string2 ):
    if sorted(string1.lower()) == sorted(string2.lower()):
        return True
    else:
        return False

However, my issue is that I wish to not return a True value should both input strings be the exact same. For example:

anagrams('silent','silent')

This outputs True but I do not want it to do that, what changes should I make to implement this?

Upvotes: 1

Views: 79

Answers (2)

Dani Mesejo
Dani Mesejo

Reputation: 61910

Just check if the strings are different:

def anagrams(string1, string2):
    if sorted(string1.lower()) == sorted(string2.lower()) and string1.lower() != string2.lower():
        return True
    else:
        return False

result = anagrams('silent', 'silent')
print(result)

Output

False

You could use Counter instead of sorted:

from collections import Counter


def anagrams(string1, string2):
    if string1.lower() != string2.lower() and Counter(string1.lower()) == Counter(string2.lower()):
        return True
    else:
        return False

print(anagrams('silent', 'silent'))
print(anagrams('silent', 'sitlen'))

Output

False
True

UPDATE

As suggested by @RoadRunner, you could do:

from collections import Counter


def anagrams(string1, string2):
    ls1, ls2 = string1.lower(), string2.lower()
    return ls1 != ls2 and Counter(ls1) == Counter(ls2)

print(anagrams('silent', 'silent'))
print(anagrams('silent', 'sitlen'))

Upvotes: 3

anubhab
anubhab

Reputation: 760

def anagrams( string1, string2 ):
  if sorted(string1.lower()) == sorted(string2.lower()) and string1.lower() != string2.lower():
    return True
  else:
    return False

print(anagrams('silent','silent'))

Upvotes: 0

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