使用Java计算字符串中单词出现次数的快速方法

I want to find number of times a word appears in a string in a fast and efficient way using Java.

我希望使用Java以快速有效的方式查找单词出现在字符串中的次数。

The words are separated by space and I am looking for complete words.

这些单词是由空格分隔的,我正在寻找完整的单词。

Example: 
string: "the colored port should be black or white or brown"
word: "or"
output: 2

for the above example, "colored" and "port" are not counted, but "or" is counted.

对于上面的例子,“有色”和“端口”不计算在内,但计算“或”。

I considered using substring() and contains() and iterating over the string. But then we need to check for the surrounding spaces which I suppose is not efficient. Also StringUtils.countMatches() is not efficient.

我考虑使用substring()和contains()并迭代字符串。但是我们需要检查周围的空间,我认为这些空间效率不高。 StringUtils.countMatches()也没有效率。

The best way I tried is splitting the string over space and iterating over the words, and then matching them against the given word:

我尝试过的最好方法是将字符串拆分到空格上并迭代单词,然后将它们与给定单词进行匹配:

String string = "the colored port should be black or white or brown";
String[] words = string.split(" ");
String word = "or";
int occurrences = 0;
for (int i=0; i<words.length; i++)
    if (words[i].equals(word))
        occurrences++;
System.out.println(occurrences);

But I am expecting some efficient way using Matcher and regex.

但我期待使用Matcher和regex的一些有效方法。

So I tested the following code:

所以我测试了以下代码:

        String string1 = "the colored port should be black or white or brown or";
        //String string2 = "the color port should be black or white or brown or";
        String word = "or";
        Pattern pattern = Pattern.compile("\\s(" + word + ")|\\s(" + word + ")|(" + word + ")\\s");
        Matcher  matcher = pattern.matcher(string1);
        //Matcher  matcher = pattern.matcher(string2);
        int count = 0;
        while (matcher.find()){
            match=matcher.group();
            count++;
        }
        System.out.println("The word \"" + word + "\" is mentioned " + count + " times.");

It is supposed to be fast enough, and gives me the right answer for string1, but not for string2 (commented). There seems to need a little change in the regex.

它应该足够快,并给我正确的string1答案,但不是string2(评论)。正则表达式似乎需要稍微改变一下。

Any ideas?

3 个解决方案

#1

public class Test {
public static void main(String[] args) {
    String str= "the colored port should be black or white or brown";
    Pattern pattern = Pattern.compile(" or ");
    Matcher  matcher = pattern.matcher(str);

    int count = 0;
    while (matcher.find())
        count++;

    System.out.println(count);    
}

}

#2

How about this? Assuming word wont have spaces.

这个怎么样?假设单词不会有空格。

string.split("\\s"+word+"\\s").length - 1;

#3

I experimented and evaluated three answers; split based and Matcher based (as mentioned in the question), and Collections.frequency() based (as mentioned in a comment above by @4castle). Each time I measured the total time in a loop repeated 10 million times. As a result, the split based answer tends to be the most efficient way:

我试验并评估了三个答案;基于分裂和基于匹配器(在问题中提到)和基于Collections.frequency()(如上面的评论中提到的@ 4castle)。每次我测量循环中的总时间重复1000万次。因此,基于拆分的答案往往是最有效的方式:

String string = "the colored port should be black or white or brown";
String[] words = string.split(" ");
String word = "or";
int occurrences = 0;
for (int i=0; i<words.length; i++)
    if (words[i].equals(word))
        occurrences++;
System.out.println(occurrences);

Then there is Collections.frequency() based answer with a little longer running time (~5% slower):

然后有基于Collections.frequency()的答案,运行时间稍长(慢约5%):

String string = "the colored port should be black or white or brown or";
String word = "or";
int count = Collections.frequency(Arrays.asList(string.split(" ")), word);
System.out.println("The word \"" + word + "\" is mentioned " + count + " times.");

The Matcher based solution (mentioned in the question) is a lot slower (~5 times more running time).

基于匹配器的解决方案(在问题中提到)要慢很多(运行时间大约多5倍)。