如何统计一定长度的字符串中的特定单词个数

Question

如何计算特定长度字符串中的特定单词。让我们考虑一个字符串——“足球是一项伟大的运动，它是全世界最受欢迎的运动，它不仅是一种运动，而且还是一个万国欢聚的节日，这也是最令人兴奋的”。字符串的总长度是145。我喜欢统计每100个字符串中有多少个'is'。让我们考虑长度为 100 的字符串的第一部分，即 - 'Football is a great game It is most popular game over the world It is not only a game but also'。在这里，我们在 100 个字符中找到了 3 个“是”。字符串的其余部分是 -'a festival of get together for the nations which is most exciting too' 其长度为 69，并且该长度中有 1 个“is”。我可以从字符串中找出给定单词的数量，但不能从特定长度的字符串中找出。下面是我的代码 -string word = "is";string sentence = "Football is a great game It is most popular game all over the world It is not only a game but also a festival of get together for the nations which is most exciting too";int count = 0;foreach (Match match in Regex.Matches(sentence, word, RegexOptions.IgnoreCase)){    count++;}Console.WriteLine("{0}" + " Found " + "{1}" + " Times", word, count);`输入：string - '足球是一项伟大的运动，它是全世界最受欢迎的运动，它不仅是一项运动，也是一个万国欢聚的节日，也是最令人兴奋的'词-'是'长度 - 100输出：在第一部分：给定单词的数量 = 3在第二部分：给定单词的数量 = 1

Answer 1

试试这个代码：

public class JavaApplication22 {

    public static void main(String[] args) {

        String str = "Football is a great game It is most popular game all over the world It is not only a game but also a festival of get together for the nations which is most exciting too";

        String pattern = "is";

        int count = 0;

        int a = 0;

        while((a = str.indexOf(pattern, a)) != -1){

            a += pattern.length();

            count++;

        }

        System.out.println("Count is : " + count);

    }

}

Answer 2

创建一个具有所需长度的子字符串，然后使用 linq 进行计数。

int length = 100;

string word = "is";

string sentence = "Football is a great game It is most popular game all over the world It is not only a game but also a festival of get together for the nations which is most exciting too";

//Substring with desired length

sentence = sentence.Substring(0, length);

int count = 0;

//creating an array of words

string[] words = sentence.Split(Convert.ToChar(" "));

//linq query

count = words.Where(x => x == word).Count();

Debug.WriteLine(count);

对于第二部分，创建一个从 100 开始到字符串末尾的子字符串。