Python challenge_7

#python #programming #webdev #challenge

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 :)

Top comments (3)

Thanh Tran • Dec 10 '22

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

Thanh Tran • Dec 10 '22

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

Phantz • Sep 27 '21 • Edited

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.

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