Question

我需要确定特定列表的频率计数。我的列表看起来像

gun,bun 
bun,gun,
pin,bin
bin,pin
stay,way.

从上面的列表中，我需要输出为

gun,bun  2
pin,bin  2
stay,way 1

请提出任何想法。

Answer 1

//this is data string
String str = "gun,bun,bun,gun,pin,bin bin,pin,stay,way";
// here we have created a map with String key and Integer values
Map<String, Integer> hm = new HashMap<String, Integer>();
//now we are iterating through each string by splitting data by "," so we'll get each string
for (String strTmp : str.split(",")) {
        //checking if map already contains the entry then update the count
        if (hm.containsKey(strTmp)) {
               Integer val = hm.get(strTmp);
               val = val + 1;
               hm.put(strTmp, val);
        } else {//else just add it
               hm.put(strTmp, 1);
        }
}
//printing the result
System.out.println(hm);

Answer 2

您还应该查看google guava library，特别是Iterables频率方法

Answer 3

使用HashMap<String,Integer>，循环显示单词列表并存储计数。请参阅此处有关如何使用HashMap：

的文档

http://download.oracle.com/javase/6/docs/api/java/util/HashMap.html

Answer 4

为每行创建一个单词列表，并使用方法Collection.containsAll（...）来检查行是否包含相同的单词集。

Answer 5

Commons Collection提供了一个“Bag”，其行为符合要求。 http://commons.apache.org/collections/apidocs/org/apache/commons/collections/Bag.html 还有一个更新的sourceforge实现可以处理泛型https://sourceforge.net/projects/collections/

Answer 6

您可以使用Collections.frequency()方法计算频率。您正在传递以逗号分隔的字符串，因此我们需要创建下面给出的此类字符串的对象。创建一个类并传递您的字符串对象（逗号分隔的String对象）并覆盖equals和hashcode方法以识别两个字符串是否等于。

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

class Test {
    public static void main(String[] args) {
        List<CommaSeperatedStringObject> list = new ArrayList<CommaSeperatedStringObject>();
        list.add(new CommaSeperatedStringObject("gun,bun"));
        list.add(new CommaSeperatedStringObject("bun,gun"));
        list.add(new CommaSeperatedStringObject("dan,dana,dan"));
        list.add(new CommaSeperatedStringObject("pin,bin"));
        list.add(new CommaSeperatedStringObject("dana,dan,dan"));
        list.add(new CommaSeperatedStringObject("bin,pin"));
        list.add(new CommaSeperatedStringObject("stay,way"));
        list.add(new CommaSeperatedStringObject("dan,dan,dana"));
        Map<CommaSeperatedStringObject, Integer> map = new HashMap<CommaSeperatedStringObject, Integer>();
        for (CommaSeperatedStringObject s : list) {
            map.put(s, Collections.frequency(list, s)); // here you can check frequency.
        }
        System.out.println(map);
    }
}

class CommaSeperatedStringObject {
    String str1;
    List<String> strList;
    public CommaSeperatedStringObject(String str) {
        this.str1 = str;
        String[] seperate = str1.split(",");
        strList = Arrays.asList(seperate);
        Collections.sort(strList);
    }

    @Override
    public boolean equals(Object obj) {
        CommaSeperatedStringObject cso = (CommaSeperatedStringObject)obj;
        boolean status = false;
        if(this.strList.equals(cso.strList)){
            status = true;
        }
        return status;
    }

    @Override
    public int hashCode() {
        return 1;
    }
    @Override
    public String toString() {
        return "\n"+strList.toString();
    }
}

使用java识别频率计数

6 个答案: