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Python - Possible English one-word anagrams with typed letters

I know that changes to this question were asked earlier, but I could not understand any of the previous implementations, because most of them are associated with the use of sets and the issubset method.

Here's what I'm trying to do: I have a set of words in a dictionary and a list of possible letters. I want to find out if the elements of a set can be formed by reinstalling the letters in the list. Here is my current implementation:

def solve(dictionary, letters):

    for word in dictionary: #for each word in the dictionary
        if len(word) > len(letters):   # first optimization, doesn't check words that are larger than letter set
            continue
        else: 
            scrambledword = "".join([b for b in sorted(list(word))]) #sorts the letters in each word
            if set(scrambledword).issubset(letters):
                print word


def main():
    dictionary = set([x.strip() for x in open("C:\\Python27\\dictionary.txt")])        
    letters = sorted(['v','r','o','o','m','a','b','c','d'])
    solve(dictionary, letters)


main()

, , "". , "" , , "a" "r". "issubset" ?

+3
3

, [oops, - ""!], , , , - . , Python . , -

from collections import Counter

letters = ['v','r','o','o','m','a','b','c','d']
words = 'cardboard boom booom'.split()
letterscount = Counter(letters)

for word in words:
    wordcount = Counter(word)
    print word, all(letterscount[c] >= wordcount[c] for c in wordcount)


cardboard False
boom True
booom False

- :

>>> c = Counter(letters)
>>> c
Counter({'o': 2, 'a': 1, 'c': 1, 'b': 1, 'd': 1, 'm': 1, 'r': 1, 'v': 1})
>>> c['o']
2
>>> c['z']
0

[DSM: ! , , Counter .]

, :

from collections import defaultdict, Counter
from itertools import combinations

# precomputations
allwords = open('/usr/share/dict/words').read().split() 
allwords = list(w for w in allwords if len(w) >= 3) # hack, /words contains lots of silliness
allwords_by_count = defaultdict(list)
for i, word in enumerate(allwords):
    allwords_by_count[frozenset(word)].append((word, Counter(word)))
    if i % 1000 == 0:
        print i, word


def wordsfrom(letters, words_by_count):
    lettercount = Counter(letters)
    for subsetsize in range(1, len(lettercount)+1):
        for subset in combinations(lettercount, subsetsize):
            for possword, posswordcount in words_by_count[frozenset(subset)]:
                if all(posswordcount[c] <= lettercount[c] for c in posswordcount):
                    yield possword

>>> wordsfrom('thistles', allwords_by_count)
<generator object wordsfrom at 0x1032956e0>
>>> list(wordsfrom('thistles', allwords_by_count))
['ess', 'sis', 'tit', 'tst', 'hei', 'hie', 'lei', 'lie', 'sie', 'sise', 'tie', 'tite', 'she', 'het', 'teth', 'the', 'els', 'less', 'elt', 'let', 'telt', 'set', 'sett', 'stet', 'test', 'his', 'hiss', 'shi', 'sish', 'hit', 'lis', 'liss', 'sil', 'lit', 'til', 'tilt', 'ist', 'its', 'sist', 'sit', 'shies', 'tithe', 'isle', 'sile', 'sisel', 'lite', 'teil', 'teli', 'tile', 'title', 'seit', 'sesti', 'site', 'stite', 'testis', 'hest', 'seth', 'lest', 'selt', 'lish', 'slish', 'hilt', 'lith', 'tilth', 'hist', 'sith', 'stith', 'this', 'list', 'silt', 'slit', 'stilt', 'liesh', 'shiel', 'lithe', 'shiest', 'sithe', 'theist', 'thesis', 'islet', 'istle', 'sistle', 'slite', 'stile', 'stilet', 'hitless', 'tehsil', 'thistle']

[. , "" , , ..]

, "" :

>>> assert all(w in allwords for w in (wordsfrom('thistles', allwords_by_count)))
>>>
+5

, , , ( ), .

. , , :

def alphabetize(word):
    "".join(sorted(word))

newDictionary , .

def solve(newDictionary, letters):
    query = alphabetize(letters)
    return query in newDictionary

: , , .

+1

collections, :

def Letters(x):
    return frozenset(Counter(x).items())

→ {anagram1, anagram2,...}:

vocabulary = ['apple', 'banana', 'rats', 'star', 'tars']
countsToWords = defaultdict(lambda: set())
for word in vocabulary:
    countsToWords[Letters(word)].add(word)

"" O (1):

def solve(query):
    return countsToWords[Letters(query)]

:

print( solve('star') )
# {'tars', 'star', 'rats'}
0
source

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