List<String> list = Arrays.asList("as", "asda", "asd", "urff", "dfkjds", "hfad", "asd", "qadasd", "as", "asda",
        "asd", "urff", "dfkjds", "hfad", "asd", "qadasd" + "as", "asda", "asd", "urff", "dfkjds", "hfad", "asd",
        "qadasd", "as", "asda", "asd", "urff", "dfkjds", "hfad", "asd", "qadasd");

方法1:

Set<String> set = new LinkedHashSet<>();
set.addAll(list);

for (String s : set) {

    System.out.println(s + " : " + Collections.frequency(list, s));
}

方法2:

int count = 1;
Map<String, Integer> map = new HashMap<>();
Set<String> set1 = new LinkedHashSet<>();
for (String s : list) {
    if (!set1.add(s)) {
        count = map.get(s) + 1;
    }
    map.put(s, count);
    count = 1;

}
System.out.println(map);

使用Java 8特性在数组中查找字符串值的简单方法。

public void checkDuplicateOccurance() {
        List<String> duplicateList = new ArrayList<String>();
        duplicateList.add("Cat");
        duplicateList.add("Dog");
        duplicateList.add("Cat");
        duplicateList.add("cow");
        duplicateList.add("Cow");
        duplicateList.add("Goat");          
        Map<String, Long> couterMap = duplicateList.stream().collect(Collectors.groupingBy(e -> e.toString(),Collectors.counting()));
        System.out.println(couterMap);
    }

输出:{猫=2，山羊=1，牛=1，牛=1，狗=1}

你可以注意到“Cow”和“Cow”不被认为是相同的字符串，如果你需要它在相同的计数下，使用. tolowercase()。请在下面找到相同的片段。

Map<String, Long> couterMap = duplicateList.stream().collect(Collectors.groupingBy(e -> e.toString().toLowerCase(),Collectors.counting()));

输出:{猫=2，牛=2，山羊=1，狗=1}

要做到这一点，有几种方法，即:

返回单个元素出现次数的方法:

采集频率

Collections.frequency(animals, "bat");

Java流:

过滤器

animals.stream().filter("bat"::equals).count();

只是迭代思考列表

public static long manually(Collection<?> c, Object o){
    int count = 0;
    for(Object e : c)
        if(e.equals(o))
            count++;
    return count;
}

创建频率映射的方法:

Collectors.groupingBy

Map<String, Long> counts = 
       animals.stream()
              .collect(Collectors.groupingBy(Function.identity(), Collectors.counting()));

走

Map<String, Long> map = new HashMap<>();
c.forEach(e -> map.merge(e, 1L, Long::sum));

手动

Map<String, Integer> mp = new HashMap<>();
        animals.forEach(animal -> mp.compute(animal, (k, v) -> (v == null) ? 1 : v + 1));

一个包含所有方法的运行示例:

import java.util.*;
import java.util.function.Function;
import java.util.stream.Collectors;

public class Frequency {

    public static int frequency(Collection<?> c, Object o){
        return Collections.frequency(c, o);
    }

    public static long filter(Collection<?> c, Object o){
        return c.stream().filter(o::equals).count();
    }

    public static long manually(Collection<?> c, Object o){
        int count = 0;
        for(Object e : c)
            if(e.equals(o))
                count++;
        return count;
    }

    public static Map<?, Long> mapGroupBy(Collection<?> c){
        return c.stream()
                .collect(Collectors.groupingBy(Function.identity() , Collectors.counting()));
    }

    public static Map<Object, Long> mapMerge(Collection<?> c){
        Map<Object, Long> map = new HashMap<>();
        c.forEach(e -> map.merge(e, 1L, Long::sum));
        return map;
    }

    public static Map<Object, Long> manualMap(Collection<?> c){
        Map<Object, Long> map = new HashMap<>();
        c.forEach(e -> map.compute(e, (k, v) -> (v == null) ? 1 : v + 1));
        return map;
    }


    public static void main(String[] args){
        List<String> animals = new ArrayList<>();
        animals.add("bat");
        animals.add("owl");
        animals.add("bat");
        animals.add("bat");

        System.out.println(frequency(animals, "bat"));
        System.out.println(filter(animals,"bat"));
        System.out.println(manually(animals,"bat"));
        mapGroupBy(animals).forEach((k, v) -> System.out.println(k + " -> "+v));
        mapMerge(animals).forEach((k, v) -> System.out.println(k + " -> "+v));
        manualMap(animals).forEach((k, v) -> System.out.println(k + " -> "+v));
    }
}

方法名称应该反映这些方法正在做什么，但是，我使用名称来反映所使用的方法(假设在当前上下文中这是可以的)。

Java 8 -另一种方法

String searched = "bat";
long n = IntStream.range(0, animals.size())
            .filter(i -> searched.equals(animals.get(i)))
            .count();

我很确定集合中的静态频率方法在这里会派上用场:

int occurrences = Collections.frequency(animals, "bat");

反正我也会这么做。我很确定这就是jdk 1.6。

package traversal;

import java.util.ArrayList;
import java.util.List;

public class Occurrance {
    static int count;

    public static void main(String[] args) {
        List<String> ls = new ArrayList<String>();
        ls.add("aa");
        ls.add("aa");
        ls.add("bb");
        ls.add("cc");
        ls.add("dd");
        ls.add("ee");
        ls.add("ee");
        ls.add("aa");
        ls.add("aa");

        for (int i = 0; i < ls.size(); i++) {
            if (ls.get(i) == "aa") {
                count = count + 1;
            }
        }
        System.out.println(count);
    }
}

输出:4