Creating an anagram function

Question

Been trying to figure this out for two hours or so, can't seem to figure out how to create an anagram function without using sorted(). Trying to figure it out without using splice, .find(), .count(), .replace(), and sorted(). Currently what I have which works:

def anagram(anag1, anag2):
    if (sorted(anag1) == sorted(anag2)):
        return True
    else:
        return False

If anybody has a way to do it without sorted that would be great.

Possible duplicate of [Checking strings against each other (Anagrams)](https://stackoverflow.com/questions/14990725/checking-strings-against-each-other-anagrams) — Thierry Lathuille, Mar 06 '19 at 05:55
@Mason Desconi, you may consider accepting the answer that helped: https://meta.stackexchange.com/questions/5234/how-does-accepting-an-answer-work cheers — DirtyBit, Mar 21 '19 at 06:20

score 2 · Answer 1 · answered Mar 06 '19 at 05:57

2

Using Counter:

from collections import Counter

def anagram(str_1, str_2):
   return Counter(str_1) == Counter(str_2)

answered Mar 06 '19 at 05:57

DirtyBit

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U13-Forward · Answer 2 · 2019-03-07T08:50:41.293

0

Try using all with a generator inside checking equality:

def anagram(anag1, anag2):
    if all(len([i for a in anag1 if a == i]) == len([i for b in anag2 if b == i]) for x in zip(anag1,anag2) for i in x):
        return True
    else:
        return False
print(anagram('car','arc'))

Output:

True

edited Mar 07 '19 at 08:50

answered Mar 06 '19 at 05:37

U13-Forward

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Sadly, I'm not able to use the sort method... If I could it would be a lot easier. – Mason Desconi Mar 06 '19 at 05:53
@MasonDesconi Edited my answer – U13-Forward Mar 06 '19 at 06:06
Ack, you got a better one than me. Just return the result of the `all`, though. – Multihunter Mar 06 '19 at 06:21
@Multihunter Lol – U13-Forward Mar 06 '19 at 06:35
@MasonDesconi Why are you unupvoting everyone – U13-Forward Mar 06 '19 at 06:35
He asked for a solution that didn't use `.count()`. It was my un-upvote. Sorry. – Multihunter Mar 06 '19 at 06:54
@Multihunter Now? – U13-Forward Mar 07 '19 at 08:50

benvc · Answer 3 · 2019-03-06T14:52:11.300

0

I like the sorted approach in your question as well as the collections.Counter approach better, but if you are just looking for alternatives you could use re.findall to determine whether each character in the first word occurs with the same frequency in the second word.

For example:

import re

def is_anagram(a, b):    
    return all(re.findall(c, a) == re.findall(c, b) for c in a)

print(is_anagram('cinema', 'iceman'))
#True

edited Mar 06 '19 at 14:52

answered Mar 06 '19 at 05:58

benvc

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Edited answer with alternative approach based on good comment from @AdamSmith that identified flaw in original approach (see the original in the edit history for context). – benvc Mar 06 '19 at 06:25

score 0 · Answer 4 · answered Mar 06 '19 at 06:12

using counter make dictionary of each input string after that we can compare those dictionaries if there is no any difference, it's anagram otherwise not.

from collections import Counter

def anagram(ang1,ang2):

    count_list1 = Counter(ang1)
    count_dict1 = dict(count_list1)

    count_list2 = Counter(ang2)
    count_dict2 = dict(count_list2)

    dict_compare_length = len(count_dict1.items() - count_dict2.items())

    if dict_compare_length==0:
        return True
    else:
        return False

print(anagram('listen','silen'))

Multihunter · Answer 5 · 2019-03-06T06:30:05.080

I assume you want to see a solution without any fancy collections or functions. Just primitive data types, loops and conditions. So, how about...

def count(word):
  map = {}
  for e in word:
    if e not in map:
      map[e] = 0
    map[e] += 1
  return map

def anagram(a, b):
  c1 = count(a)
  c2 = count(b)
  return all(c1[k] == c2[k] for k in c1)

Make a dictionary that lists the number of each element, then compares that there's the same number in both.

Creating an anagram function

5 Answers5