Trying to come up with python anagram function

Question

What I\'m trying to do is if I have a list like:

[\"lime\", \"mile\", \"liem\", \"tag\", \"gat\", \"goat\", \"math\"]

I want to write a functio

Answer 1

The easy way to analyses anagram words is to put them in alphabetic order.So you create a second list with alphabetic ordered words.

['lime', 'mile', 'liem', 'tag', 'gat']

index = 0
b = []
for i in a:
    b.insert(index, ''.join(sorted(i)))
    index = index + 1

['eilm', 'eilm', 'eilm', 'agt', 'agt']

I think you can have more pythonesque code than the one i give you, but i think the important thing for you is to order letters in the word.

Now you can do something to analyse your anagrams

Answer 2

def does_string_contain(big_word, small_word) :
    list_string = list(big_word)
    for char in small_word:
        if char in list_string:
            list_string.pop(list_string.index(char))
        else:
            return False
    for char in small_word:
        if char in list_string:
            return False
    return True

Answer 3

Checks two given strings if they're anagrams or not. Those strings may include spaces, numbers or special characters

#First of all define a function that counts the number of alphabets in a string. It'll be used as a final condition to check for anagrams
def count_alpha(text):
    text = text.lower()
    count = 0
    i = 97    #ASCII code range from 'a' to 'z' is 97 to 122
    while i < 123:
        if chr(i) in text:
            count += text.count(chr(i))
        i += 1
    return count
text1 = input('Enter your First Word: ')
text2 = input('Enter your Second Word: ')
#replace all the spaces with empty strings and make the string lower case 
text1 = text1.replace(' ','').lower()
text2 = text2.replace(' ','').lower()
i = 97
while i < 123:
    #check if an alphabet count in both strings is the same.
    if text1.count(chr(i)) == text2.count(chr(i)):
        #replace all the alphabets with spaces
        text1 = text1.replace(chr(i),' ')
        text2 = text2.replace(chr(i),' ')
    i += 1  
#since all the alphabets have been replaced by spaces. There's no alphabet left(if they had the same number of particular alphabets)
if count_alpha(text1) == 0 and count_alpha(text2) == 0:
    print('They are anagrams')
else: print('They are not anagrams')

So here's your code. Enjoy!

Answer 4

You can use itertools to create all permutations of the words, remove the word you just found the permutations of, and then check your list one word at a time to see if it is in permutations like so

from itertools import permutations

l = ["lime", "mile", "liem", "tag", "gat", "goat", "math"]
final = []
perms = []
for i in l:
    perms += [''.join(p) for p in permutations(i)]
    perms.remove(i)

for i in l:
    if i in perms:
        final.append(i)
print final

This isn't the fastest solution in the world, especially if you have long words like 'resistance', 'ancestries'

Answer 5

an approach identifying the words by the frozenset of the characters:

from collections import defaultdict

wordlist = ["lime", "mile", "liem", "tag", "gat", "goat", "math"]

worddict = defaultdict(list) 
for word in wordlist:
    worddict[frozenset(word)].append(word)

anagrams = [words for words in worddict.values() if len(words) > 1]
print(anagrams)

# [['lime', 'mile', 'liem'], ['tag', 'gat']]

the output is not yet quite what you wanted, but flattening that list is easy if you prefer that.

---

update after comments:

the solution above will not handle words with repeating characters well. but this will (this time the key of the dictionary is just the string consisting of the sorted letters):

for word in wordlist:
    worddict[''.join(sorted(word))].append(word)

Answer 6

An algorithm to check if two words are anagrams in python.

1) Take two words: e.g.

("mile", "lime") ("tiles", "miles")

2) Make a string array/list:

(['m', 'i', 'l', 'e'], ['l', 'i', 'm', 'e']) (['t', 'i', 'l','e', 's'], ['m', 'i', 'l', 'e', 's'])

3) Sort arrays

(['e', 'i', 'l', 'm'], ['e', 'i', 'l', 'm']) (['e', 'i', 'l','s', 't'], ['e', 'i', 'l', 'm', 's'])

4)Check if first_array[i] == second_array[i] for 0<=i<=len(first_array)||second_array

5) Conclusion. If 4) is held, return true, else false.

from itertools import combinations

def anagram(w1,w2):
    list1 = list(w1)
    list2 = list(w2)

    list1.sort()
    list2.sort()

    idx = 0
    is_anagram = True

    while idx < len(w1) and is_anagram:
        if list1[idx]== list2[idx]:
            idx += 1
        else:
            is_anagram = False
    return is_anagram


lst_words = ["lime", "mile", "liem", "tag", "gat", "goat", "math"]
lst_anagrams = set()
for i in combinations(lst_words, 2):
    if anagram(*i):
        lst_anagrams |= set(i) 

print list(lst_anagrams)