Python-给定输入字母的可能的英语单字变位词

Question

我知道之前有人问过这个问题的变体，但我无法理解之前的任何实现，因为它们中的大多数都涉及使用集合和 issubset 方法。

这是我正在尝试做的事情:我在字典中有一组单词和一个可能的字母列表。我想看看是否可以通过重新排列列表中的字母来形成集合的成员。这是我当前的实现:

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()

此实现的一个明显问题是会发现某些词在“字母”中使用多个字母。例如，单词“cardboard”显示为有效单词，尽管字母列表中只有一个副本“a”和“r”。如何在列表上使用“issubset”方法？

Answer 1

要知道您是否可以用一组字母组成单词 [哎呀，我自己做的 - 我的意思是“集合”!]，您希望每个字母至少出现正确的次数，所以我认为我们将不得不以某种方式在那里工作。根据定义，Python 集不关心源列表中元素的数量。也许像

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)

给予

cardboard False
boom True
booom False

Counter 是一个方便的实用程序类:

>>> 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

[帝斯曼:回归!我删除了一个无效的社区编辑，因为 Counter 实例不可哈希。]

如果搜索速度是一个问题，那么您可以权衡内存和预计算时间:

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']

[呵呵。我只是注意到“thiSTLes”本身不在列表中，但那是因为它不在单词文件中..]

是的，单词文件中确实存在明显的“非单词”:

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