从StringArray java中的标记/单词构建句子

Question

我如何将list<String>中的单词连接起来，使其回到句子的真实形式

String sentence = "i get money. i love it. i want to buy. something. ";
String[] arrSent = sentence.split("\\. ");

for(int i=0; i<arrSent.length; i++) {
    String[] words = arrSent[i].split("\\ ");
    for(int j=0; j<words.length; j++ {
        listWord.add(words[j]);
    }
}

输出为：

i
get
money
i
love
it
i
want
to
buy
something

我只是想把它重建成真实的形式（如句子）

更新

我试过像你建议的那样。但我找到了一条新的困难之路

我已经从列表中删除了一个单词“爱”，并将其添加到新列表“listWord2”。当我把它重建成句子的真实形式时，新句子中的.就消失了

代码如下：

String [] arrays2 = listKata2.toArray(new String[listWord2.size()]);
sentence = Arrays.deepToString(arrays2).replaceAll(",", "");
System.out.println("result :  : "+sentence);

输出为：

[i get money i it i want to buy something]

缺少.

我应该再次按空格分割listWord2吗？，请推荐我

Answer 1

做：

String sentence = "i get money. i love it. i want to buy. something. ";
String[] arrSent = sentence.split("\\. ");

String sentenceRebuild = "";
for(int i=0; i<arrSent.length; i++){
  String[] words = arrSent[i].split("\\ ");

  for(int j=0; j<words.length; j++){
    sentenceRebuild += words[j];
  }
}

Answer 2

尝试以下内容：

String sentence = "i get money. i love it. i want to buy. something. ";
String[] arrSent = sentence.split(" ");
sentence=Arrays.toString(arrSent).replaceAll(",", "");
System.out.println(sentence);

输出：

[i get money. i love it. i want to buy. something.]

上面是如果你想在句子中.，下面是没有.

String sentence = "i get money. i love it. i want to buy. something. ";
String[] arrSent = sentence.split("\\. ");
sentence=Arrays.toString(arrSent).replaceAll(",", "");
System.out.println(sentence);

输出：

[i get money i love it i want to buy something]

Answer 3

唯一真正的答案是，一旦你把所有的单词都记下来，失去了句子之间的句号，就没有办法把它们带回来——这些信息将永远丢失，然后就不可能重建原始结构

你需要弄清楚如何（如果，实际上是）保留这些信息。一种方法是保留句子数组，只需用单词列表而不是句子字符串填充它，类似这样：

List<List<String>> sentences = new List<List<String>>();
String[] arrSent = sentence.split("\\. ");
for (int i = 0; i < arrSent.length; i++)
    sentences.add(Arrays.asList(arrSend[i].split("\\ "));

然后你会得到

(
   ( "i", "get", "money" ),
   ( "i", "love", "it" ),
   ( "i", "want", "to", "buy" ),
   ( "something" )
)

这很容易看出如何从中重建原始文本

另一种选择可能是保持单词列表的扁平化，但为句子终止的位置添加特殊的占位符，例如使用null值。扫描单词的算法应该知道如何在不崩溃的情况下处理这些占位符，而重建句子的算法将使用这些占位符来添加句号：

String[] arrSent = sentence.split("\\. ");

for(int i=0; i<arrSent.length; i++) {
    String[] words = arrSent[i].split("\\ ");
    for(int j=0; j<words.length; j++ {
        listWord.add(words[j]);
    }
    listWord.add(null);
}

// Rebuild
StringBuffer output = new StringBuffer();
for (Iterator<String> it = listWord.iterator(); it.hasNext(); ) {
     String val = it.next();
     String nextword = (output.length() > 0 ? " " : "") + val;
     output.append(val == null ? "." : nextword);
}