从Java中的另一个字符串中删除字符串 [英] Removing strings from another string in java

问题描述

让我们说我有这个单词列表:

Lets say I have this list of words:

 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"

是否存在与stopWord匹配并在忽略大小写时将其删除的方法；像这样的地方吗?:

Is there method that matches the stopWords and removes them while ignoring case; like this somewhere out there?:

 String noStopWordsText = remove(text, stopWords);

结果:

 " would like do nice novel nature people"

如果您了解正则表达式，效果很好，但是我真的更喜欢像Commons解决方案这样的东西，它更注重性能.

If you know about regex that wold work great but I would really prefer something like commons solution that is bit more performance oriented.

顺便说一句，现在我正在使用以下通用方法，该方法缺乏适当的不区分大小写的处理方式:

BTW, right now I'm using this commons method which is lacking proper insensitive case handling:

 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);     

推荐答案

这是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，因为它更长且不清楚，但是如果性能确实非常重要，那么这就是 O(n)，其中 n 是文字的长度.

This is a solution that does not use regular expressions. I think it's inferior to my other answer because it is much longer and less clear, but if performance is really, really important then this is O(n) where n is the length of the text.

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());

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