从字典中查找句子字谜的有效方法？

Question

我需要编写一个程序，该程序将带有字典和任意字符串的文件作为输入，然后输出该字典中构成给定字符串的字谜的所有单词组合。 例如，使用英语中最流行的 100 个单词和字符串 "i not work"，我应该得到类似 [' on it work', ' into work', ' not i work', ' know or it', ' work it no', ' to work in'] 的内容。

问题是我的程序效率太低了：字典中有 100 个单词，字符串长度的实际限制是 7 个字符，之后的所有内容都需要很长时间。我尝试寻找与此事相关的各种算法均无济于事。

这是我搜索字谜的方法：

def sortstring(string):
    return ''.join(sorted(string))

def simplify(all_strings):
    possible_strings = defaultdict(list)
    for string in all_strings:
        possible_strings[sortstring(string).strip()].append(string)
    return possible_strings

def generate(database, length,curstring="", curdata=set()):
    if len(curstring.replace(" ", "")) > length:
        return set()
    if len((curstring).replace(" ", "")) == length:
        return curdata.union(set([curstring]))
    for i in database:
        if len((curstring+i).replace(" ", "")) <= length:
            curdata = curdata.union(generate(database.difference(set([i])),
                length, curstring+" "+i, curdata))
            database = database.difference(set([i]))
    return curdata

def analyse(database, input_string):
    cletters = countstring(input_string)
    strings = simplify(generate(database, cletters))
    data = list()
    sorted_string = sortstring(input_string).strip()
    if sorted_string in strings.keys():
        data = strings[sorted_string]
    return len(strings.values()), data

def countstring(string):
    a = countletters(string)
    return sum(a.values())

def countletters(string):
    result = {}
    for i in ascii_lowercase:
        result[i] = string.count(i)
    return result

谁能提出改进的方法？尽管我认为我使用的算法应该被完全抛弃，因为它的复杂性似乎太高了，因为它太慢了。 以防万一：该程序应该足够高效以支持数万个单词和多达数十个字符的字符串的字典。这比我做的要好得多。

Answer 1

我自己解决了部分问题。 解决了生成器代码中的 for-if 反模式：

def generate(database, length,letters,curstring="",curdata=set()):
if len(curstring.replace(" ",""))>length:
    return set()
if len((curstring).replace(" ",""))==length:
    return curdata.union(set([curstring]))
t=countletters(curstring)
for i in ascii_lowercase:
    if t[i]>letters[i]:
        return set()
for i in database:
    t=countletters(curstring+i)
    test=0
    for j in ascii_lowercase:
        if t[j]>letters[j]:
            test=1
    if test: continue
    if sum(t.values())<=length:
        curdata=curdata.union(generate(database.difference(set([i])),length,letters,curstring+" "+i,curdata))
        database=database.difference(set([i]))
return curdata

现在速度快了很多，但如果字典包含数万个单词和/或输入字符串很长，它仍然很慢。

Answer 2

这是实现我在 cmets 中建议的树方法的递归方法：

def frequencyDict(s):
    s = s.lower()
    d = {}
    for c in s:
        if c.isalpha():
            if c in d:
                d[c] += 1
            else:
                d[c] = 1
    return d

def canMake(w,fdict):
    d = frequencyDict(w)
    return all(d[c] <= fdict.get(c,0) for c in d)

def candidates(wlist,fdict):
    return [w for w in wlist if canMake(w,fdict)]

def anagrams(wlist,fdict):
    if len(wlist) == 0 or len(fdict) == 0:
        return "no anagrams"
    hits = []
    firstWords = candidates(wlist,fdict)
    if len(firstWords) == 0:
        return "no anagrams"
    for w in firstWords:
        #create reduced frequency dict
        d = fdict.copy() 
        for c in w:
            d[c] -= 1
            if d[c] == 0: del d[c]
        #if d is empty, the first word is also a the last word
        if len(d) == 0:
            hits.append(w)
        else:
            #create reduced word list
            rlist = [v for v in wlist if canMake(v,d)]
            tails = anagrams(rlist, d)
            if tails != "no anagrams":
                hits.extend(w + " " + t for t in tails)
    if len(hits) == 0:
        return "no anagrams"
    else:
        return hits

def findAnagrams(wlist,s):
    return anagrams(wlist,frequencyDict(s.lower()))

f = open("linuxwords.txt")
words = f.read().split('\n')
f.close()
words = [w.strip().lower() for w in words if not '-' in w]
test = findAnagrams(words, "Donald Trump")

从旧的 Linux 单词列表中找出所有 730 个“Donald Trump”字谜大约需要 20 秒。我最喜欢的是“干货领主”