最有效的方法 - 测试 2 个字谜 Python 的字符串

Question

There're many ways of testing if strings are anagrams . However, I wonder if there is a way to iterate over each word only once? And if not what's the most efficient way to do it in Python?

We can traverse through the second string checking whether each character is present in the first string. However that gives us n-1 iterations over the first string at worst case scenario (reversgram), when using build-in __contains__() method ( __iter__() method is called).

def is_anagram(str_1, str_2):
  #chceck if same length
  if (len(str_1) != len(str_2)):
    return False
  else:
    #lowercase all characters  
    str1, str2 = list(str_1.lower()),list(str_2.lower())
    for letter in str1:
      if letter not in str2:
        return False
      str2.remove(letter)
    return True

Is there any other way?

Answer 1

if you can use Collections.Counter then it becomes simple, because if two words are anagrams they would have same keys and same values.

from collections import Counter
def is_anagram(word1,word2):
    return Counter(word1)==Counter(word2)

word1 = 'ahbgrettf'
word2 = 'arethbfgt'

print(is_anagram(word1,word2)

to add on to @Maxime's answer if we use defaultdict we dont have to check if a key exists then check if keys match and values match to decide if its an anagram.

from collections import defaultdict

def is_anagram(word1,word2):
    table1, table2 = defaultdict(int), defaultdict(int)

    for c in word1:
        table1[c]+=1

    for c in word2:
        table2[c]+=1

    if set(table1.keys()) == set(table2.keys()):
        for k, v in table1.items():
            if table2[k]!=v:
                return False
    else:
        return False
    return True

print(is_anagram('ahbgrettf','arethbfgt'))

Answer 2

Maybe with dictionaries?

edit: added dan's suggestion

word1 = 'ahbgrettf'
word2 = 'arethbfgt'


def is_anagram(word1, word2):

    if (len(word1) != len(word2)):
        return False

    word_dic = {}

    # n iterations
    for char in word1:
        if word_dic.get(char):
            word_dic[char] += 1
        else:
            word_dic[char] = 1

    # n iterations
    for char in word2:
        if word_dic.get(char):
            word_dic[char] -= 1
        else:
            return False

    # n iterations
    for v in word_dic.values():
        if v != 0:
            return False

    return True


print(is_anagram(word1, word2))

total: 3n?

Answer 3

You can use defaultdict to have a default value and create a dictionary of letter frequencies and subtract from it with the other string with O(3n)

from collections import defaultdict

def is_anagram2(str_1, str_2):
    #check if same length
    if (len(str_1) != len(str_2)):
        return False
    #creates a dictionary with default value of 0 for all keys
    str_1_dict = defaultdict(int)

    #adds how many of each letter in the dictionary
    for i in str_1:
        str_1_dict[i] += 1
    #subracts how many of each letter in the dictionary
    for i in str_2:
        str_1_dict[i] -= 1
    #checks to make sure all values are 0 (same number of each letter in both strings)
    for i in str_1_dict:
        if not str_1_dict[i] == 0:
            return False
    return True
is_anagram2('aaaa','aaaa')

Answer 4

I think using a dictionary is indeed the fastest since sorting takes at least O(nlogn). Creating dictionaries on the other hand should take O(n + n) or O(n) effectively. The.get() ensures if the key is not already there, return default 0 and then add 1 to insert the key and initialize the value to 1 in the dictionary. At the end, equating the two dictionaries makes sure the the same key:value pairs exist in both dictionaries. Optionally, you can check the length of two string and return a false in the beginning if the length does not match.

def anagram_checker(str1, str2):
    str1 = str1.replace(" ", "").lower() #optional
    str2 = str2.replace(" ", "").lower() #optional

    str1_char_dict = {}
    str2_char_dict = {}

    for char in str1:
        str1_char_dict[char] = str1_char_dict.get(char, 0) + 1

    for char in str2:
        str2_char_dict[char] = str2_char_dict.get(char, 0) + 1

    return str1_char_dict == str2_char_dict