CODEWITHSUNDEEP
pythonanagram
colalove
colalove

Reputation: 31

How to write a function that judges if two strings are anagrams in Python?

The input of this function is two strings, and the aim is to judge whether these two strings are anagrams. For example, "qwerty" and "qetyrw" are anagrams, which means rearrange the characters in first string can get the second string, case-insensitively. "qwerty" and "QWerTY" are also anagrams. I'm just confused with my function, which returns nothing.

My function is as follows:

def is_anagram(string_a,string_b):
    """returns True if the strings are anagrams of each other

    str, list -> boolean"""
    new_a=string_a.lower()
    new_b=string_b.lower()
    i=0
    if len(string_a)!=len(string_b):
        return False
    else:
        while i<=len(new_a)-1:
            if new_a[i] in new_b:
                list(new_b).remove(new_a[i])
            i=i+1
            break
        if len(list(new_b))==0:
            return True
        else:
            return False

Upvotes: 0

Views: 1397

Answers (4)

JuniorCompressor
JuniorCompressor

Reputation: 20015

Sorting to identify if a string is anagram of another, takes O(n logn) time. Instead, we can use a counter object that counts the occurrences of each letter:

from collections import Counter

def is_anagram(string_a, string_b):
    return Counter(string_a.lower()) == Counter(string_b.lower())

This will take on average linear time. The basic observation is that two anagram strings have exactly the same letter frequencies.

Examples:

>>> is_anagram("abc", "aaa")
False
>>> is_anagram("abc", "aac")
False
>>> is_anagram("abc", "acb")
True

Upvotes: 5

xirururu
xirururu

Reputation: 5508

Your code is wrong by "list(new_b)", it is not stored, so every time it will be regarded as a new list.

And the "break" in the while loop is also not necessary.

Please try the following code:

def is_anagram(string_a,string_b):
    """returns True if the strings are anagrams of each other

    str, list -> boolean"""
    new_a=string_a.lower()
    new_b=string_b.lower()
    i=0
    if len(string_a)!=len(string_b):
        return False
    else:
        list_b = list(new_b) # store the list(new_b) into list_b
        while i<=len(new_a)-1:
            if new_a[i] in list_b:
                list_b.remove(new_a[i])
            i=i+1
           #break should be removed! 
        if len(list_b)==0:
            return True
        else:
            return False

Upvotes: 0

shx2
shx2

Reputation: 64298

You can use the Counter class:

from collections import Counter

def is_anagram(string_a,string_b):
    return Counter(string_a.lower()) == Counter(string_b.lower())

Upvotes: 1

Julien Spronck
Julien Spronck

Reputation: 15423

You just have to sort the strings:

def is_anagram(string_a, string_b):
    return sorted(string_a.lower()) == sorted(string_b.lower())

Upvotes: 3

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