Python challenge_7

Anagrams

level of challenge 3/10

• Two strings are anagrams
• if you can make one from the other by rearranging the letters.
• Write a function
• named is_anagram that takes two strings as its parameters.
• Your function should return True if the strings are anagrams,

• and False otherwise.

• For example,

• the call is_anagram("typhoon", "opython") should return True while the call

• is_anagram("Alice", "Bob") should return False.

Hint

• You can compare how many times each letter appears in each string.
• Alternatively, sorting the letters in each string makes this much easier.

My solution

def is_anagram(string_one, string_two):
    str_dic_one = {}
    str_dic_tow = {}

    for str_one in string_one:
        if str_one in str_dic_one:
            str_dic_one[str_one] += 1
        else:
            str_dic_one[str_one] = 1

    for str_tow in string_two:
        if str_tow in str_dic_tow:
            str_dic_tow[str_tow] += 1
        else:
            str_dic_tow[str_tow] = 1

    if str_dic_one == str_dic_tow:
        return True
    else:
        return False

print(is_anagram('typhoon', 'opython'))

Another solution

def is_anagram(string1, string2):
    return sorted(string1) == sorted(string2)

def count_letters(string):
    return {l: string.count(l) for l in string}

def is_anagram(string1, string2):
    return count_letters(string1) == count_letters(string2)

Add your solution in the comment :)

Comments

[Thanh Tran]

def is_anagram(word1,word2):
    if sorted(word1)== sorted(word2):
        return True
    else:
        return False

[Thanh Tran]

def is_anagram(word1,word2):
if sorted(word1)== sorted(word2):
return True
else:
return False

[Phantz]

from collections import Counter

is_anagram = lambda x, y: Counter(x) == Counter(y)

I'm gonna guess Counter would be able to do it pretty fast, if not the fastest.