My anagram code doesn't account for multiple letters

Question

So my code gets two words, and checks if one is the anagram of another one.

However doesn't work if multiple letters are exchanged, although I tried to account for that.

storedword = input("Enter your primary word \t")
global word 
word = list(storedword)


word3 = input("Enter anagram word \t")
word3lowercase = word3.lower()
anaw = list(word3lowercase)


counter = int(0)
letterchecker = int(0)
listlength = len(word)
newcounter = int(0)
if len(anaw) != len(word):
    print ("not anagram")

if len(anaw) == len(word):
    while counter < listlength and newcounter < listlength:
        tempcount = 0
        if anaw[counter] == word[newcounter]:
            temp = word[newcounter]
            word[newcounter] = word[tempcount]
            word[tempcount]=temp
            letterchecker +=1
            counter +=1
            tempcount +=1
            newcounter = int(0)

        else:
            newcounter +=1

if counter == len(word):
    print ("anagram")
else:
    print ("not anagram")

I think it's gone somewhere wrong after the if len(anaw) section, for example if the primary word is "hannah", and the secondary word is "hannnn", it thinks it's an anagram.

Can you explain the logic you tried to implement here? Also... `int(0)`??? — Idos, Mar 01 '16 at 15:18
Also you can simply go `sorted(first_word) == sorted(second_word)` and save yourself all this trouble — Idos, Mar 01 '16 at 15:20
ah, not allowed to use sorted fuction, had to be loops and stuff — I Haque, Mar 01 '16 at 15:22
@IHaque If an answer has solved your question please [accept it](http://meta.stackexchange.com/a/5235). Thank you — Idos, Mar 02 '16 at 19:22

Idos · Answer 1 · 2016-03-01T15:36:35.500

3

There is much simpler logic that can be implemented here, even without using sorted and such. Let's assume you have a function anagram:

def anagram(word1, word2):
    if len(word1) != len(word2):
        return False

    def char_count(word):
        char_count = {}
        for c in word:
            char_count[c] = char_count.get(c, 0) + 1
        return char_count

    cr1 = char_count(word1)
    cr2 = char_count(word2)
    return cr1 == cr2

You can test this with:

>>> print(anagram("anagram", "aanragm"))
True
>>> print(anagram("anagram", "aangtfragm"))
False

And for future readers, a super simple pythonic solution might be using Counter:

from collections import Counter
>>> Counter(word1) == Counter(word2)

Or using sorted:

>>> sorted(word1) == sorted(word2)

edited Mar 01 '16 at 15:36

answered Mar 01 '16 at 15:26

Idos

15,053
14
60
75

You can use a `set` to avoid re-checking similar characters. Also, I'd add the `Counter(word) == Counter(otherword)` solution for the sake of future readers. – Reut Sharabani Mar 01 '16 at 15:29
1

@ReutSharabani I figured he might not be able to import/use `Counter`. Regarding the `set` I might edit later (not next to a computer so I don't want to make any changes without testing them). You are welcome to edit my answer tho ;) – Idos Mar 01 '16 at 15:31
I know he isn't, but the answer doesn't have to be too specific. If a future reader runs into this he may want the nicer solution. That's why I wrote "for the sake of *future* readers" :-) – Reut Sharabani Mar 01 '16 at 15:32
if it is for future readers, you should precise `from collections import Counter` – mvelay Mar 01 '16 at 15:33
1

Haha okay sure I'll edit this a lil later then for all the millions of future readers who will flock to this answer like wild geese ;p – Idos Mar 01 '16 at 15:33
For the other answers that aren't restricted. [Determine if 2 lists have the same elements, regardless of order?](http://stackoverflow.com/q/8866652/5012922) – SirParselot Mar 01 '16 at 15:49

Jan · Accepted Answer · 2016-03-01T16:04:26.083

0

newcounter = int(0)

This is the line that causes the trouble (in the while loop). Because of it you start checking the word from the beginning again. I think you want it to be newcounter=letterchecker. Since already used characters are put to the front of word they are ignored if you start with letterchecker

Tell me if it works

Edit:Checked with example given, seems to work.

edited Mar 01 '16 at 16:04

answered Mar 01 '16 at 15:27

Jan

1,504
10
15

sure `int(0)` causes me trouble too – mvelay Mar 01 '16 at 15:28
Nah, the same problem occurs – I Haque Mar 01 '16 at 15:31
Did you change the first one, or the one in the while loop? I meant the one in the loop. – Jan Mar 01 '16 at 15:34

Alex · Answer 3 · 2016-03-01T15:41:03.003

0

Without using sort you could use the following approach. It removes a letter from an array of a characters of the second word. The words are only anagrams if there are no letters left (and the words are the same length to start with and have a length larger than zero):

word1="hannah"
word2="nahpan"

chars1= list(word1)
chars2= list(word2)

if len(chars1)==len(chars2) and len(chars1)>0:
    for char in chars1:
        if char not in chars2:
           break
        chars2.remove(char)                   


if len(chars2)==0:
    print "Both words are anagrams"
else:
    print "Words are not anagrams"

edited Mar 01 '16 at 15:41

answered Mar 01 '16 at 15:37

Alex

21,273
10
61
73

what does that char for char mean? – I Haque Mar 01 '16 at 15:39
It's just all the chars in each word, it will make `chars1` to be `['h', 'a', 'n', 'n', 'a', 'h']` – Idos Mar 01 '16 at 15:39
It converts the words in an array of characters. I have changed the code to use `list`, looks a bit cleaner. – Alex Mar 01 '16 at 15:40

user1723662 · Answer 4 · 2016-03-01T23:32:51.013

-2

[EDIT THIS IS JUST FOR PALINDROMES I CANT READ]

Here is something a bit more simple:

storedword = input("Enter your primary word \t")
word3 = input("Enter anagram word \t")
if storedword == word3[::-1]:
    print "Is Anagram"
else:
    print "Is not anagram"

edited Mar 01 '16 at 23:32

answered Mar 01 '16 at 15:27

user1723662

29
4

1

Anagrams are not palindromes. – Reut Sharabani Mar 01 '16 at 15:28
1

but palindromes are anagrams :) – mvelay Mar 01 '16 at 15:29
@user1723662: this is wrong, `word3[::-1]` simply reverts `words3` – mvelay Mar 01 '16 at 15:50
Sorry when I saw anagram I though palindrome! – user1723662 Mar 01 '16 at 23:32
@user1723662 you do know that you can **delete** your answer, right? – Idos Mar 02 '16 at 19:21

My anagram code doesn't account for multiple letters

4 Answers4