Ruby 无法组合字母中的英文单词

Question

我需要找到所有可以由字符串中的字母组成的英文单词

 sentence="Ziegler's Giant Bar"

我可以制作一个字母数组

 sentence.split(//)  

如何从 Ruby 中的句子中生成超过 4500 个英文单词？

[编辑]

最好将问题分成几部分：

1. 只创建一个不超过 10 个字母的单词数组
2. 较长的单词可以单独查找

Answer 1

[假设您可以在一个单词中重复使用源字母]：对于字典列表中的每个单词，构造两个字母数组 - 一个用于候选单词，一个用于输入字符串。从单词 array-of-letters 中减去输入的字母 array，如果没有剩下任何字母，那么你就有了匹配项。执行此操作的代码如下所示：

def findWordsWithReplacement(sentence)
    out=[]
    splitArray=sentence.downcase.split(//)
    `cat /usr/share/dict/words`.each{|word|
        if (word.strip!.downcase.split(//) - splitArray).empty?
            out.push word
        end
     }
     return out
end

您可以像这样从 irb 调试器调用该函数：

output=findWordsWithReplacement("some input string"); puts output.join(" ")

...或者这是一个包装器，您可以使用它来从脚本交互地调用函数：

puts "enter the text."
ARGF.each {|line|
    puts "working..."
    out=findWordsWithReplacement(line)
    puts out.join(" ")
    puts "there were #{out.size} words."
}

在 Mac 上运行时，输出如下所示：

$ ./findwords.rb
输入文字。
齐格勒的巨型酒吧
工作...
啊啊啊 aal aalii Aani Ab aba abaiser abalienate Abantes Abaris abas abase abaser Abasgi abasia Abassin abate abater abatis abaze abb 阿爸 阿巴斯·阿巴斯·阿巴斯·阿巴斯·阿巴斯 Abbie Abe abear 阿贝尔阿贝勒 阿贝利亚 阿贝尔阿贝尔阿贝尔特里阿贝尔特里 Aberia aberrant aberrate 教唆教唆 Abie Abies abietate abietene abietin Abietineae Abiezer Abigail abigeat abilla abintestate
[....]
Z Z za Zabaean zabeta 扎比安 zabra zabti zabtie zag zain Zan zanella zant zante Zanzalian zanze 桑给巴尔 zar zaratite zareba zat zati zattare Zea zeal 无心 无心 斑马 斑马 Zebrina zebrine zee zein zeist zel Zelanian Zeltinger Zen Zenaga zenana zer zest zeta 齐亚拉齐亚拉特齐贝林 zibet ziega zieger zig zigzag zigzagger Zilla zing zingel Zingiber Zingiberene Zinnia zinsang Zinzar zira zirai 齐尔巴尼特 Zirian Zirianian Zizania Zizia zizz
共有 6725 个单词。

这远远超过 4500 个单词，但那是因为 Mac 词典非常大。如果您想准确重现 Knuth 的结果，请从此处下载并解压缩 Knuth 的字典：http://www.packetstormsecurity.org/Crackers/wordlists/dictionaries/knuth_words.gz 并将“/usr/share/dict/words”替换为您解压缩替代目录的路径。如果你做对了，你会得到 4514 个单词，以这个集合结尾：

zanier zanies zaniness 桑给巴尔 zazen zeal zebra zebras Zeiss zeitgeist Zen Zennist zest zestier zeta 齐格勒 zig 曲折曲折曲折曲折曲折曲折曲折曲折曲折曲折曲折曲折曲折曲曲折曲折曲折曲折曲折曲折曲折曲折曲折曲折曲折曲折曲折曲折曲折曲曲 zingier zings 百日草

我相信这回答了最初的问题。

另外，提问者/阅读者可能想要列出可以从字符串构造的所有单词而不重复使用任何输入字母。我建议的代码实现如下：复制候选词，然后对于输入字符串中的每个字母，从副本中破坏性地删除该字母的第一个实例（使用“切片！”）。如果此过程吸收了所有字母，则接受该单词。

def findWordsNoReplacement(sentence)
    out=[]
    splitInput=sentence.downcase.split(//)
    `cat /usr/share/dict/words`.each{|word|
        copy=word.strip!.downcase
        splitInput.each {|o| copy.slice!(o) }
        out.push word if copy==""
     }
     return out
end

Answer 2

如果您想查找其字母和频率受给定短语限制的单词， 您可以构建一个正则表达式来为您执行此操作：

sentence = "Ziegler's Giant Bar"

# count how many times each letter occurs in the 
# sentence (ignoring case, and removing non-letters)
counts = Hash.new(0)
sentence.downcase.gsub(/[^a-z]/,'').split(//).each do |letter|
  counts[letter] += 1
end
letters = counts.keys.join
length = counts.values.inject { |a,b| a + b }

# construct a regex that matches upto that many occurences
# of only those letters, ignoring non-letters
# (in a positive look ahead)
length_regex = /(?=^(?:[^a-z]*[#{letters}]){1,#{length}}[^a-z]*$)/i
# construct regexes that matches each letter up to its
# proper frequency (in a positive look ahead)
count_regexes = counts.map do |letter, count|
  /(?=^(?:[^#{letter}]*#{letter}){0,#{count}}[^#{letter}]*$)/i
end

# combine the regexes, to form a regex that will only
# match words that are made of a subset of the letters in the string
regex = /#{length_regex}#{count_regexes.join('')}/

# open a big file of words, and find all the ones that match
words = File.open("/usr/share/dict/words") do |f|
  f.map { |word| word.chomp }.find_all { |word| regex =~ word }
end

words.length #=> 3182
words #=> ["A", "a", "aa", "aal", "aalii", "Aani", "Ab", "aba", "abaiser", "Abantes",
          "Abaris", "abas", "abase", "abaser", "Abasgi", "abate", "abater", "abatis",
          ...
          "ba", "baa", "Baal", "baal", "Baalist", "Baalite", "Baalize", "baar", "bae",
          "Baeria", "baetzner", "bag", "baga", "bagani", "bagatine", "bagel", "bagganet",
          ...
          "eager", "eagle", "eaglet", "eagre", "ean", "ear", "earing", "earl", "earlet",
          "earn", "earner", "earnest", "earring", "eartab", "ease", "easel", "easer",
          ...
          "gab", "Gabe", "gabi", "gable", "gablet", "Gabriel", "Gael", "gaen", "gaet",
          "gag", "gagate", "gage", "gageable", "gagee", "gageite", "gager", "Gaia",
          ...
          "Iberian", "Iberis", "iberite", "ibis", "Ibsenite", "ie", "Ierne", "Igara",
          "Igbira", "ignatia", "ignite", "igniter", "Ila", "ilesite", "ilia", "Ilian",
          ...
          "laang", "lab", "Laban", "labia", "labiate", "labis", "labra", "labret", "laet",
          "laeti", "lag", "lagan", "lagen", "lagena", "lager", "laggar", "laggen",
          ...
          "Nabal", "Nabalite", "nabla", "nable", "nabs", "nae", "naegate", "naegates",
          "nael", "nag", "Naga", "naga", "Nagari", "nagger", "naggle", "nagster", "Naias",
          ...
          "Rab", "rab", "rabat", "rabatine", "Rabi", "rabies", "rabinet", "rag", "raga",
          "rage", "rager", "raggee", "ragger", "raggil", "raggle", "raging", "raglan",
          ...
          "sa", "saa", "Saan", "sab", "Saba", "Sabal", "Saban", "sabe", "saber",
          "saberleg", "Sabia", "Sabian", "Sabina", "sabina", "Sabine", "sabine", "Sabir",
          ...
          "tabes", "Tabira", "tabla", "table", "tabler", "tables", "tabling", "Tabriz",
          "tae", "tael", "taen", "taenia", "taenial", "tag", "Tagabilis", "Tagal",
          ...
          "zest", "zeta", "ziara", "ziarat", "zibeline", "zibet", "ziega", "zieger",
          "zig", "zing", "zingel", "Zingiber", "zira", "zirai", "Zirbanit", "Zirian"]

Positive lookaheads 让您可以创建一个匹配字符串中某个指定模式匹配的 位置 的正则表达式，而不会消耗匹配的字符串部分。 我们在这里使用它们将相同的字符串与单个正则表达式中的多个模式进行匹配。 仅当我们的所有模式都匹配时，该位置才匹配。

如果我们允许无限重复使用原始短语中的字母（就像 Knuth 根据glenra 的评论所做的那样），那么构造正则表达式就更容易了：

sentence = "Ziegler's Giant Bar"

# find all the letters in the sentence
letters = sentence.downcase.gsub(/[^a-z]/,'').split(//).uniq

# construct a regex that matches any line in which
# the only letters used are the ones in the sentence
regex = /^([^a-z]|[#{letters.join}])*$/i

# open a big file of words, and find all the ones that match
words = File.open("/usr/share/dict/words") do |f|
  f.map { |word| word.chomp }.find_all { |word| regex =~ word }
end

words.length #=> 6725
words #=> ["A", "a", "aa", "aal", "aalii", "Aani", "Ab", "aba", "abaiser", "abalienate",
           ...
           "azine", "B", "b", "ba", "baa", "Baal", "baal", "Baalist", "Baalite",
           "Baalize", "baar", "Bab", "baba", "babai", "Babbie", "Babbitt", "babbitt",
           ...
           "Britannian", "britten", "brittle", "brittleness", "brittling", "Briza",
           "brizz", "E", "e", "ea", "eager", "eagerness", "eagle", "eagless", "eaglet",
           "eagre", "ean", "ear", "earing", "earl", "earless", "earlet", "earliness",
           ...
           "eternalize", "eternalness", "eternize", "etesian", "etna", "Etnean", "Etta",
           "Ettarre", "ettle", "ezba", "Ezra", "G", "g", "Ga", "ga", "gab", "gabber",
           "gabble", "gabbler", "Gabe", "gabelle", "gabeller", "gabgab", "gabi", "gable",
           ...
           "grittiness", "grittle", "Grizel", "Grizzel", "grizzle", "grizzler", "grr",
           "I", "i", "iba", "Iban", "Ibanag", "Iberes", "Iberi", "Iberia", "Iberian",
           ...
           "itinerarian", "itinerate", "its", "Itza", "Izar", "izar", "izle", "iztle",
           "L", "l", "la", "laager", "laang", "lab", "Laban", "labara", "labba", "labber",
           ...
           "litter", "litterer", "little", "littleness", "littling", "littress", "litz",
           "Liz", "Lizzie", "Llanberisslate", "N", "n", "na", "naa", "Naassenes", "nab",
           "Nabal", "Nabalite", "Nabataean", "Nabatean", "nabber", "nabla", "nable",
           ...
           "niter", "nitraniline", "nitrate", "nitratine", "Nitrian", "nitrile",
           "nitrite", "nitter", "R", "r", "ra", "Rab", "rab", "rabanna", "rabat",
           "rabatine", "rabatte", "rabbanist", "rabbanite", "rabbet", "rabbeting",
           ...
           "riteless", "ritelessness", "ritling", "rittingerite", "rizzar", "rizzle", "S",
           "s", "sa", "saa", "Saan", "sab", "Saba", "Sabaean", "sabaigrass", "Sabaist",
           ...
           "strigine", "string", "stringene", "stringent", "stringentness", "stringer",
           "stringiness", "stringing", "stringless", "strit", "T", "t", "ta", "taa",
           "Taal", "taar", "Tab", "tab", "tabaret", "tabbarea", "tabber", "tabbinet",
           ...
           "tsessebe", "tsetse", "tsia", "tsine", "tst", "tzaritza", "Tzental", "Z", "z",
           "za", "Zabaean", "zabeta", "Zabian", "zabra", "zabti", "zabtie", "zag", "zain",
           ...
           "Zirian", "Zirianian", "Zizania", "Zizia", "zizz"]

Answer 3

我认为 Ruby 没有英文词典。但是您可以尝试将原始字符串的所有排列存储在一个数组中，然后对照 Google 检查这些字符串？说一个词实际上是一个词，如果有超过 100.000 次点击或什么？

Answer 4

你可以像这样得到一个字母数组：

sentence = "Ziegler's Giant Bar"
letters = sentence.split(//)