如何在长随机字符串中找到可能的英语单词？

Question

I'm doing an artistic project where I want to see if any information emerges from a long string of characters (~28,000). It's sort of like the problem one faces in solving a Jumble. Here's a snippet:

What's the most efficient way of searching for all possible English words embedded (both forwards and backwards) in this string?

What is a useful dictionary against which to check the substrings? Is there a good library for doing this sort of thing? I have searched around and found some interesting TRIE solutions; but most of them are dealing with the situation where you know the set of words in advance.

Answer 1

I used this solution to find all words forwards and backwards from a corpus of 28,000 random characters in a dictionary of 100,000 words in .5 seconds. It runs in O(n) time. It takes a file called "words.txt" which is a dictionary that has words separated by some kind of whitespace. I used the default unix wordlist in /usr/share/dict/words but I'm sure you can find plenty of text file dictionaries online if not that one.

from random import choice
import string

dictionary = set(open('words.txt','r').read().lower().split())
max_len = max(map(len, dictionary)) #longest word in the set of words

text = ''.join([choice(string.ascii_lowercase) for i in xrange(28000)])
text += '-'+text[::-1] #append the reverse of the text to itself

words_found = set() #set of words found, starts empty
for i in xrange(len(text)): #for each possible starting position in the corpus
    chunk = text[i:i+max_len+1] #chunk that is the size of the longest word
    for j in xrange(1,len(chunk)+1): #loop to check each possible subchunk
        word = chunk[:j] #subchunk
        if word in dictionary: #constant time hash lookup if it's in dictionary
            words_found.add(word) #add to set of words

print words_found

Answer 2

Here is a bisection/binary search that should be usefull.

def isaprefix(frag, wordlist, first, last):
    """
    Recursive binary search of wordlist for words that start with frag.

    assumes wordlist is a sorted list
    typically called with first = 0 and last = len(wordlist)

    first,last -->> integer
    returns bool
    """

    # base case - down to two elements
    if (last - first) < 2:
        # return False unless frag is a prefix
        # of either of the two remaining words
        return wordlist[first].startswith(frag) or wordlist[last].startswith(frag)

    #mid = (first + last)/2
    midword = wordlist[(first + last) / 2]

    # go ahead and return if you find one
    # a second base case?
    if midword.startswith(frag):
        return True

    #print word, ' - ', wordlist[mid], ' - ', wordlist[mid][:len(word)], ' - ', isprefix
    # start the tests
    # python does just fine comparing strings
    if frag < midword:
        # set the limits to the lower half
        # of the previous range searched and recurse
        return isaprefix(frag, wordlist, first, mid-1)

    # frag is > midword: set the limits to the upper half
    # of the previous range searched and recurse
    return isaprefix(frag, wordlist, mid+1, last)

Answer 3

您可以考虑从整个字典中创建一个序列，然后将它们对齐以使用史密斯水人或任何启发式局部对齐算法获取序列中的单词