Python-给出输入字母可能的英文单字字谜

我知道以前曾经问过这个变种,但我无法理解以前的任何实现,因为大多数都涉及使用集合和issubset方法.

这是我想要做的：我在字典中有一组单词和可能的字母列表.我想找到是否可以通过重新排列列表中的字母来形成集合的成员.这是我目前的实施：

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()

这个实现的明显问题是会发现在“字母”中使用多个字母的单词.例如,尽管字母列表中只有一个“a”和“r”副本,但“纸板”这个词显示为有效单词.如何在列表中使用“issubset”方法？

解决方法:

要知道你是否可以从一组字母中找出一个字[哎呀,我自己做了 – 我的意思是’收集’！],你希望每封信至少发生一次,所以我想我们’不得不以某种方式在那里工作.根据定义,Python集合不关心源列表中的元素数量.也许是这样的

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

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实例不可清除.

如果关注搜索速度,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']

[嘿.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,明显的“非词”实际上在文件中：

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

标签：membership,anagram,python,list,set
来源： https://codeday.me/bug/20190826/1735276.html