从java中的另一个字符串中删除字符串

Question

可以说我有这个单词列表:

 String[] stopWords = new String[]{"i","a","and","about","an","are","as","at","be","by","com","for","from","how","in","is","it","not","of","on","or","that","the","this","to","was","what","when","where","who","will","with","the","www"};

比我有文字

 String text = "I would like to do a nice novel about nature AND people"

是否有匹配stopWords的方法并在忽略大小写的情况下删除它们; 像这样在某个地方？:

 String noStopWordsText = remove(text, stopWords);

结果:

 " would like do nice novel nature people"

如果你知道正则表达式工作得很好但我真的更喜欢像公共解决方案更具有性能导向的东西.

顺便说一句,现在我正在使用这种缺乏适当的不敏感案例处理的公共方法:

 private static final String[] stopWords = new String[]{"i", "a", "and", "about", "an", "are", "as", "at", "be", "by", "com", "for", "from", "how", "in", "is", "it", "not", "of", "on", "or", "that", "the", "this", "to", "was", "what", "when", "where", "who", "will", "with", "the", "www", "I", "A", "AND", "ABOUT", "AN", "ARE", "AS", "AT", "BE", "BY", "COM", "FOR", "FROM", "HOW", "IN", "IS", "IT", "NOT", "OF", "ON", "OR", "THAT", "THE", "THIS", "TO", "WAS", "WHAT", "WHEN", "WHERE", "WHO", "WILL", "WITH", "THE", "WWW"};
 private static final String[] blanksForStopWords = new String[]{"", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", "", ""};

 noStopWordsText = StringUtils.replaceEach(text, stopWords, blanksForStopWords);     

Answer 1

使用停用词创建正则表达式,使其不区分大小写,然后使用匹配器的replaceAll方法将所有匹配替换为空字符串

import java.util.regex.*;

Pattern stopWords = Pattern.compile("\\b(?:i|a|and|about|an|are|...)\\b\\s*", Pattern.CASE_INSENSITIVE);
Matcher matcher = stopWords.matcher("I would like to do a nice novel about nature AND people");
String clean = matcher.replaceAll("");

将...在模式只是我懒惰,继续停止词列表.

另一种方法是循环对所有的停用词和使用String的replaceAll方法.这种方法的问题是replaceAll将为每个调用编译一个新的正则表达式,因此在循环中使用它并不是非常有效.此外,你不能传递,使正则表达式不区分大小写,当您使用该标志String的replaceAll.

编辑:我添加\b了模式,使其只匹配整个单词.我还添加\s*了使其在任何空格之后的全球化,这可能没有必要.

Answer 2

您可以创建一个reg表达式以匹配所有停用词 [例如a,在此处注释空格]并最终得到

str.replaceAll(regexpression,"");

要么

 String[] stopWords = new String[]{" i ", " a ", " and ", " about ", " an ", " are ", " as ", " at ", " be ", " by ", " com ", " for ", " from ", " how ", " in ", " is ", " it ", " not ", " of ", " on ", " or ", " that ", " the ", " this ", " to ", " was ", " what ", " when ", " where ", " who ", " will ", " with ", " the ", " www "};
        String text = " I would like to do a nice novel about nature AND people ";

        for (String stopword : stopWords) {
            text = text.replaceAll("(?i)"+stopword, " ");
        }
        System.out.println(text);

输出:

 would like do nice novel nature people 

• IdeOneDemo

可能有更好的方法.

Answer 3

这是一个不使用正则表达式的解决方案。我认为它不如我的其他答案，因为它更长且不太清晰，但如果性能真的非常重要，那么这是O(n)，其中n是文本的长度。

Set<String> stopWords = new HashSet<String>();
stopWords.add("a");
stopWords.add("and");
// and so on ...

String sampleText = "I would like to do a nice novel about nature AND people";
StringBuffer clean = new StringBuffer();
int index = 0;

while (index < sampleText.length) {
  // the only word delimiter supported is space, if you want other
  // delimiters you have to do a series of indexOf calls and see which
  // one gives the smallest index, or use regex
  int nextIndex = sampleText.indexOf(" ", index);
  if (nextIndex == -1) {
    nextIndex = sampleText.length - 1;
  }
  String word = sampleText.substring(index, nextIndex);
  if (!stopWords.contains(word.toLowerCase())) {
    clean.append(word);
    if (nextIndex < sampleText.length) {
      // this adds the word delimiter, e.g. the following space
      clean.append(sampleText.substring(nextIndex, nextIndex + 1)); 
    }
  }
  index = nextIndex + 1;
}

System.out.println("Stop words removed: " + clean.toString());