Python strings with anagrams

Question

At the moment this code takes in a string from a user and compares it to a text file in which many words are stored. It then outputs all the words that contain an exact match to the string. (E.G "otp = opt, top, pot) Currently when i input the string it only matches the string to the word with the EXACT same letters in a rearranged order.

My question is how do i go about being able to type in excess letters but still output all the words that are contained? for example: Type in "orkignwer" and the program will output "working" even though there are extra letters.

words = []


def isAnAnagram(word, user):
    wordList= list(word)
    wordList.sort()
    inputList= list(user)
    inputList.sort()
    return (wordList == inputList)

def getAnagrams(user):
    lister = [word for word in words if len(word) == len(user) ]
    for item in lister:
        if isAnAnagram(item, user):
            yield item


with open('Dictionary.txt', 'r') as f:
    allwords = f.readlines()
f.close()

for x in allwords:
    x = x.rstrip()
    words.append(x)
inp = 1


while inp != "99":
    inp = input("enter word:")
    result = getAnagrams(inp)
    print(list(result))

You will likely want to use a Counter and then check that the input word has all the same letters (a.keys() == b.keys()) and that each letter has higher or equal counts b[k] > v for k, v in a.items(). — Alex Huszagh
– Alex Huszagh, Commented Mar 8, 2016 at 23:48
Aside: remember there's no need to close the file when using with. — cdonts
– cdonts, Commented Mar 8, 2016 at 23:48

bbkglb · Accepted Answer · 2016-03-09 04:54:05Z

1

You have to edit the isAnAnagram and the getAnagrams functions. First the getAnagrams function should be edited to also include the words of greater length in the lister list:

def getAnagrams(user):
    lister = [word for word in words if len(word) <= len(user) ]
    for item in lister:
        if isAnAnagram(item, user):
            yield item

Then you would need to edit the isAnAnagram function. As Alexander Huszagh pointed out, you can use the Counter from the collections package:

from collections import Counter

def isAnAnagram(word, user):
    word_counter = Counter(word)
    input_counter = Counter(user)
    return all(count <= input_counter[key] for key, count in word_counter.items())

The all(count <= input_counter[key] for key, count in word_counter.items()) checks to see if every letter of word appears in user at least as many times as they did in word.

P.S. If you want a more optimized solution, you might want to checkout TRIEs (e.g. MARISA-trie, python-trie or PyTrie).

edited Mar 9, 2016 at 4:54

answered Mar 9, 2016 at 0:32

bbkglb

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2 Comments

CBren Over a year ago

Thanks everyone. I used your code changes bbkglb but it still does not output what i need. What i need is, for example, if i typed in "tpozxc" the prog would realise that there are 3 words in this and output "opt, top ,pot" but what is happening is it is ONLY outputting strings that match exactly in a re-arranged order, so to get top opt and pot i need to enter in a string with ONLY a combination of those letters.

bbkglb Over a year ago

@CBren Oh I see, I misunderstood your question then; I edited both of the functions. This should fix it.

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