Python – 对乱码字中的字进行高效狩猎

我想你可以将其分类为拼字游戏风格的问题,但是由于朋友提到英国电视测验节目倒数计时,它开始了.演出中的各种表演包括参赛者被提出了一组乱码的信件,他们必须提出最长的话.我的朋友提到的是“RAEPKWAEN”.

在相当短的时间内,我用Python鞭打了一些东西来处理这个问题,使用PyEnchant来处理字典查找,但是我注意到它真的不能扩展这些.

这是我目前所在的

#!/usr/bin/python

from itertools import permutations
import enchant
from sys import argv

def find_longest(origin):
    s = enchant.Dict("en_US")
    for i in range(len(origin),-1):
        print "Checking against words of length %d" % i
        pool = permutations(origin,i)
        for comb in pool:
            word = ''.join(comb)
            if s.check(word):
                return word
    return ""

if (__name__)== '__main__':
    result = find_longest(argv[1])
    print result

就像他们在节目中使用的9个字母的例子,9个阶乘= 362,880和8个阶乘= 40,320就好了.在这个规模上,即使它必须检查所有可能的排列和字长度,这不是很多.

然而,一旦你达到了14个字符,这是87,178,291,200个可能的组合,这意味着你依赖运气,迅速找到一个14个字符的字.

用上面的例子,我的机器大约需要12 1/2秒才能找到“reawaken”.使用14个字符的加密字,我们可以在23天的时间内谈话,只是为了检查所有可能的14个字符的排列.

有没有更有效的方法来处理这个？

解决方法

from collections import defaultdict,Counter


def find_longest(origin,known_words):
    return iter_longest(origin,known_words).next()

def iter_longest(origin,known_words,min_length=1):
    origin_map = Counter(origin)
    for i in xrange(len(origin) + 1,min_length - 1,-1):
        for word in known_words[i]:
            if check_same_letters(origin_map,word):
               yield word

def check_same_letters(origin_map,word):
    new_map = Counter(word)
    return all(new_map[let] <= origin_map[let] for let in word)

def load_words_from(file_path):
    known_words =  defaultdict(list)
    with open(file_path) as f:
        for line in f:
            word = line.strip()
            known_words[len(word)].append(word)
    return known_words

if __name__ == '__main__':
    known_words = load_words_from('words_list.txt')
    origin = 'raepkwaen'
    big_origin = 'raepkwaenaqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnmqwertyuiopasdfghjklzxcvbnm'
    print find_longest(big_origin,known_words)
    print list(iter_longest(origin,5))

输出(对于我的小58000字dict)：

counterrevolutionaries
['reawaken','awaken','enwrap','weaken','weaker','apnea','arena','awake','aware','newer','paean','parka','pekan','prank','prawn','preen','renew','waken','wreak']

笔记：

>这是简单的实现,没有优化.
> words_list.txt – 可以在Linux上使用/usr/share / dict / words.

UPDATE

如果我们只需要找到一次词,并且我们有字典,按长度排序的单词,例如通过这个脚本：

with open('words_list.txt') as f:
    words = f.readlines()
with open('words_by_len.txt','w') as f:
    for word in sorted(words,key=lambda w: len(w),reverse=True):
        f.write(word)

我们可以找到最长的字,而不是加载完整的dict到内存：

from collections import Counter
import sys


def check_same_letters(origin_map,word):
    new_map = Counter(word)
    return all(new_map[let] <= origin_map[let] for let in word)

def iter_longest_from_file(origin,file_path,min_length=1):
    origin_map = Counter(origin)
    origin_len = len(origin)
    with open(file_path) as f:
        for line in f:
            word = line.strip()
            if len(word) > origin_len:
                continue
            if len(word) < min_length:
                return
            if check_same_letters(origin_map,word):
                yield word

def find_longest_from_file(origin,file_path):
    return iter_longest_from_file(origin,file_path).next()

if __name__ == '__main__':
    origin = sys.argv[1] if len(sys.argv) > 1 else 'abcdefghijklmnopqrstuvwxyz'
    print find_longest_from_file(origin,'words_by_len.txt')

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