使用Ruby在源文本中查找词典单词

使用Ruby，我需要输出字典中的单词列表，可以通过从源文本中删除字母来形成

例如，如果我输入源文本crazed，我想得到的不仅仅是像craze和razed这样的单词，它们的字母顺序相同，并且在源文本中彼此相邻，而且还有像rad和red这样的单词，因为这些单词是存在的，并且可以通过从craized中删除select字母来找到，并且输出的单词保持字母顺序。但是，像dare或race这样的单词不应该出现在输出列表中，因为dare或race中的字母顺序与crazed中的字母顺序不同。如果raed或crae是词典中的单词，它们将成为输出的一部分

我的想法是以二进制的方式浏览源文本

(for "crazed", we'd get: 
000001 = "d"; 
000010 = "e"; 
000011 = "ed"; 
000100 = "z"; 
000101 = "zd"; 
000111 = "zed"; 
001000 = "a"; 
001001 = "ad"; etc.) 

并将每个结果与字典中的单词进行比较，尽管我不知道如何编码，也不知道这是否是最有效的。在这里，我将从你的帮助中受益匪浅

而且，源文本的长度是可变的；它不一定像《疯狂》那样有六个字母长。输入可能会更大20-30个字符，可能更多

---

我在这里搜索了一下，发现了一些关于字谜和任何字母顺序的单词的问题，但并没有具体说明我在寻找什么。这在Ruby中可能吗？谢谢。

首先，让我们将字典中的单词读入数组，然后对重复的单词进行筛选、降格和删除，例如，如果字典同时包含a和a，就像我在下面使用的Mac上的字典一样

DICTIONARY = File.readlines("/usr/share/dict/words").map { |w| w.chomp.downcase }.uniq
  #=> ["a", "aa", "aal", "aalii",..., "zyzomys", "zyzzogeton"]  
DICTIONARY.size
  #=> 234371 

下面的方法根据顺序生成给定单词的一个或多个字符的所有组合，并针对每个组合，将字符合并成字符串，检查字符串是否在字典中，如果在字典中，则将字符串保存到数组中

为了检查字符串是否与字典中的单词匹配，我使用以下方法执行二进制搜索。这利用了字典已经按字母顺序排序的事实

def subwords(word)
  arr = word.chars
  (1..word.size).each.with_object([]) do |n,a|
    arr.combination(n).each do |comb|
      w = comb.join
      a << w if DICTIONARY.bsearch { |dw| w <=> dw }
    end
  end
end

subwords "crazed"
  # => ["c", "r", "a", "z", "e", "d",
  #     "ca", "ce", "ra", "re", "ae", "ad", "ed",
  #     "cad", "rad", "red", "zed",
  #     "raze", "craze", "crazed"] 

---

下面是一个广泛的解决方案集，其中包括可以通过以任何顺序使用字母获得的单词。使用组合查找可能的子词的缺点是缺少组合的排列。根据“重要性”，在某个时候会出现“mpa”的组合。由于这不是字典中的单词，因此将跳过它。因此，我们损失了排列“地图”——字典中“重要性”的子词。下面是一个广泛的解决方案，可以找到更多可能的字典单词。我同意我的方法可以优化速度

#steps
#split string at ''
#find combinations for n=2 all the way to n=word.size
#for each combination
#find the permutations of all the arrangements
#then
#join the array
#check to see if word is in dictionary
#and it's not already collected
#if it is, add to collecting array

require 'set'
Dictionary=File.readlines('dictionary.txt').map(&:chomp).to_set
Dictionary.size #39501

def subwords(word)
    #split string at ''
  arr=word.split('')

  #excluding single letter words
  #you can change 2 to 1 in line below to select for single letter words too
  (2..word.size).each_with_object([]) do |n,a|

    #find combinations for n=2 all the way to n=word.size
    arr.combination(n).each do |comb|

        #for each combination
        #find the permutations of all the arrangements
      comb.permutation(n).each do |perm|
        
        #join the array
        w=perm.join

        #check to see if word is in dictionary and it's not already collected
        if Dictionary.include?(w) && !a.include?(w)

            #if it is, add to collecting array
          a<<w
        end
      end
    end
  end
end

p subwords('crazed')
#["car", "arc", "rec", "ace", "cad", "are", "era", "ear", "rad", "red", "adz", "zed", "czar", "care", "race", "acre", "card", "dace", "raze", "read", "dare", "dear", "adze", "daze", "craze", "cadre", "cedar", "crazed"]
p subwords('battle')
#["bat", "tab", "alb", "lab", "bet", "tat", "ate", "tea", "eat", "eta", "ale", "lea", "let", "bate", "beat", "beta", "abet", "bale", "able", "belt", "teat", "tale", "teal", "late", "bleat", "table", "latte", "battle", "tablet"]

---

第一个问题是你在用谁的字典？我不明白你为什么要把这些转换成二进制，除非你有一本方便的二进制到英语字典，里面有一个你可以使用word的ruby字符串单词。每个字符都会给你一个字符数组。另请看：@whodini9，word.chars就够了。我刚刚注意到我说我在回答中使用了bsearch，但在我的代码中找不到它。我修好了。

#steps
#split string at ''
#find combinations for n=2 all the way to n=word.size
#for each combination
#find the permutations of all the arrangements
#then
#join the array
#check to see if word is in dictionary
#and it's not already collected
#if it is, add to collecting array

require 'set'
Dictionary=File.readlines('dictionary.txt').map(&:chomp).to_set
Dictionary.size #39501

def subwords(word)
    #split string at ''
  arr=word.split('')

  #excluding single letter words
  #you can change 2 to 1 in line below to select for single letter words too
  (2..word.size).each_with_object([]) do |n,a|

    #find combinations for n=2 all the way to n=word.size
    arr.combination(n).each do |comb|

        #for each combination
        #find the permutations of all the arrangements
      comb.permutation(n).each do |perm|
        
        #join the array
        w=perm.join

        #check to see if word is in dictionary and it's not already collected
        if Dictionary.include?(w) && !a.include?(w)

            #if it is, add to collecting array
          a<<w
        end
      end
    end
  end
end

p subwords('crazed')
#["car", "arc", "rec", "ace", "cad", "are", "era", "ear", "rad", "red", "adz", "zed", "czar", "care", "race", "acre", "card", "dace", "raze", "read", "dare", "dear", "adze", "daze", "craze", "cadre", "cedar", "crazed"]
p subwords('battle')
#["bat", "tab", "alb", "lab", "bet", "tat", "ate", "tea", "eat", "eta", "ale", "lea", "let", "bate", "beat", "beta", "abet", "bale", "able", "belt", "teat", "tale", "teal", "late", "bleat", "table", "latte", "battle", "tablet"]