Linq用句子來說話

Question

我有一個單詞列表和一個句子列表。 我想知道哪些話可以在句子中找到。

這是我的代碼：

List<string> sentences = new List<string>();
List<string> words = new List<string>();

sentences.Add("Gallia est omnis divisa in partes tres, quarum unam incolunt Belgae, aliam Aquitani, tertiam qui ipsorum lingua Celtae, nostra Galli appellantur.");
sentences.Add("Alea iacta est.");
sentences.Add("Libenter homines id, quod volunt, credunt.");

words.Add("est");
words.Add("homines");

List<string> myResults = sentences
  .Where(sentence => words
     .Any(word => sentence.Contains(word)))
  .ToList();

我需要的是一個元組列表。 句子和單詞，在句子中找到。

Answer 1

首先，我們必須定義什么是單詞 。 讓它成為字母和撇號的任意組合 。

  Regex regex = new Regex(@"[\p{L}']+");

其次，我們應該考慮如何處理案件 。 讓我們實現不區分大小寫的例程：

  HashSet<string> wordsToFind = new HashSet<string>(StringComparer.OrdinalIgnoreCase) {
    "est",
    "homines"
  };

然后我們可以使用正則Regex匹配句子中的單詞，並使用Linq查詢句子：

碼：

  var actualWords = sentences
    .Select((text, index) => new {
      text = text,
      index = index,
      words = regex
        .Matches(text)
        .Cast<Match>()
        .Select(match => match.Value)
        .ToArray()
    })
    .SelectMany(item => item.words
       .Where(word => wordsToFind.Contains(word))
       .Select(word => Tuple.Create(word, item.index + 1)));

  string report = string.Join(Environment.NewLine, actualWords);

  Console.Write(report);

結果：

  (est, 1)         // est appears in the 1st sentence
  (est, 2)         // est appears in the 2nd sentence as well
  (homines, 3)     // homines appears in the 3d sentence

如果你想要Tuple<string, string> for word ， sentence ，只需在最后一個Select更改Tuple.Create(word, item.index + 1) Tuple.Create(word, item.text) Tuple.Create(word, item.index + 1)

Answer 2

你的意思是：

IEnumerable<(string, string)> query =
    from sentence in sentences
    from word in words
    where sentence.Contains(word)
    select (sentence, word);

這給了：

Answer 3

你可以這樣試試，

var result = from sentence in sentences
             from word in words
             where sentence.Contains(word)
             select Tuple.Create(sentence, word);