Find a word that all characters match other words in python

Question

Like umbellar = umbrella, both words are equal.

Login = ["umbrella", "goa", "umbrella", "back", "aery", "alem", "ayre", "gnu", "eyra", "egma", "game" "leam", "amel", "year", "meal", "yare", "gun", "alme", ​​"ung", "male", "lame", "mela", "mage"]

therefore, the output should be:

exit = [["Umbellar", "umbrella"], ["Back", "goa"], ["Brood", "Air", "eyra", "yar", "year"], ["Alem", “Alme”, “Amel”, “Lame”, “Lim”, “man”, “food”, “Mela”], [“Wildebeest”, “gun”, “Un”] [“EGMA”, “game” , "magician"],]

Answer 1

from itertools import groupby

def group_words(word_list):
    sorted_words = sorted(word_list, key=sorted)
    grouped_words = groupby(sorted_words, sorted)
    for key, words in grouped_words:
        group = list(words)
        if len(group) > 1:
            yield group

Example:

>>> group_words(["umbellar","goa","umbrella","ago","aery","alem","ayre","gnu","eyra","egma","game","leam","amel","year","meal","yare","gun","alme","ung","male","lame","mela","mage" ])
<generator object group_words at 0x0297B5F8>
>>> list(_)
[['umbellar', 'umbrella'], ['egma', 'game', 'mage'], ['alem', 'leam', 'amel', 'meal', 'alme', 'male', 'lame', 'mela'], ['aery', 'ayre', 'eyra', 'year', 'yare'], ['goa', 'ago'], ['gnu', 'gun', 'ung']]

Answer 2

They are not equal words, they are anagrams.

Anagrams can be found by sorting by character:

sorted('umbellar') == sorted('umbrella')

Answer 3

collections.defaultdict :

from collections import defaultdict

input = ["umbellar","goa","umbrella","ago","aery","alem","ayre","gnu",
"eyra","egma","game","leam","amel","year","meal","yare","gun",
"alme","ung","male","lame","mela","mage" ]

D = defaultdict(list)
for i in input:
    key = ''.join(sorted(input))
    D[key].append(i)

output = D.values()

[['umbellar', 'umbrella'], ['goa', 'ago'], ['gnu', 'gun', 'ung'], ['alem', 'leam', 'amel', 'meal', 'alme', 'male', 'lame', 'mela'], ['egma', 'game', 'mage'], ['aery', 'ayre', 'eyra', 'year', 'yare']]

Answer 4

, . . ( ) , , , , , .

input = ["umbellar","goa","umbrella","ago","aery","alem","ayre","gnu",
"eyra","egma","game","leam","amel","year","meal","yare","gun",
"alme","ung","male","lame","mela","mage" ]
res = dict()
for word in input:
    res[word]=[word]
for word in input:
    #the len test is just to avoid sorting and comparing words of different len
    candidates = filter(lambda x: len(x) == len(word) and\
                                  sorted(x) == sorted(word),res.keys())
    if len(candidates):
        canonical = candidates[0]
        for c in candidates[1:]:
            #we delete all candidates expect the canonical/
            del res[c]
            #we add the others to the canonical member
            res[canonical].append(c)
print res.values()

algth...

[['year', 'ayre', 'aery', 'yare', 'eyra'], ['umbellar', 'umbrella'],
 ['lame', 'leam', 'mela', 'amel', 'alme', 'alem', 'male', 'meal'],
 ['goa', 'ago'], ['game', 'mage', 'egma'], ['gnu', 'gun', 'ung']]

Answer 5

... , .... 'groupby()'....... ..... ....

def group_words(word_list):
    global new_list
    list1 = [] 
    _list0 = []
    _list1 = []
    new_list = []
    for elm in word_list:
        list_elm = list(elm)
        list1.append(list(list_elm))
    for ee in list1:
        ee = sorted(ee)
        ee = ''.join(ee)
        _list1.append(ee)   
    _list1 = list(set(_list1))
    for _e1 in _list1:
        for e0 in word_list:
            if  len(e0) == len(_e1):
                list_e0 = ''.join(sorted(e0))
                if _e1 == list_e0:
                    _list0.append(e0)
                    _list0 = list(_list0)
        new_list.append(_list0)
        _list0 = []
    return new_list

[['umbellar', 'umbrella'], ['goa', 'ago'], ['gnu', 'gun', 'ung'], ['alem', 'leam', 'amel', 'meal', 'alme', 'male', 'lame', 'mela'], ['egma', 'game', 'mage'], ['aery', 'ayre', 'eyra', 'year', 'yare']]