Java 8 –如何对地图排序-蒲公英云

Java 8 –如何对地图排序

电玩女神 2023-02-15 05:07 91阅读 0赞

Java 8 Stream示例，用于通过键或值对Map进行排序。

1.快速说明

在Java 8中对地图进行排序的步骤。

将地图转换成流
解决

收集并返回一个新的LinkedHashMap （保留订单）

Map result = map.entrySet().stream()

.sorted(Map.Entry.comparingByKey())             
.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,
(oldValue, newValue) -> oldValue, LinkedHashMap::new));

PS默认情况下， Collectors.toMap将返回一个HashMap

2.按键排序

SortByKeyExample.java

package com.mkyong.test;
import java.util.HashMap;
import java.util.LinkedHashMap;
import java.util.Map;
import java.util.stream.Collectors;
public class SortByKeyExample {
    public static void main(String[] argv) {
        Map<String, Integer> unsortMap = new HashMap<>();
        unsortMap.put("z", 10);
        unsortMap.put("b", 5);
        unsortMap.put("a", 6);
        unsortMap.put("c", 20);
        unsortMap.put("d", 1);
        unsortMap.put("e", 7);
        unsortMap.put("y", 8);
        unsortMap.put("n", 99);
        unsortMap.put("g", 50);
        unsortMap.put("m", 2);
        unsortMap.put("f", 9);
        System.out.println("Original...");
        System.out.println(unsortMap);
        // sort by keys, a,b,c..., and return a new LinkedHashMap
        // toMap() will returns HashMap by default, we need LinkedHashMap to keep the order.
        Map<String, Integer> result = unsortMap.entrySet().stream()
                .sorted(Map.Entry.comparingByKey())
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,
                        (oldValue, newValue) -> oldValue, LinkedHashMap::new));
        // Not Recommend, but it works.
        //Alternative way to sort a Map by keys, and put it into the "result" map
        Map<String, Integer> result2 = new LinkedHashMap<>();
        unsortMap.entrySet().stream()
                .sorted(Map.Entry.comparingByKey())
                .forEachOrdered(x -> result2.put(x.getKey(), x.getValue()));
        System.out.println("Sorted...");
        System.out.println(result);
        System.out.println(result2);
    }
}

输出量

Original...
{a=6, b=5, c=20, d=1, e=7, f=9, g=50, y=8, z=10, m=2, n=99}
Sorted...
{a=6, b=5, c=20, d=1, e=7, f=9, g=50, m=2, n=99, y=8, z=10}
{a=6, b=5, c=20, d=1, e=7, f=9, g=50, m=2, n=99, y=8, z=10}

3.按值排序

SortByValueExample.java

package com.mkyong.test;
package com.mkyong;
import java.util.Comparator;
import java.util.HashMap;
import java.util.LinkedHashMap;
import java.util.Map;
import java.util.stream.Collectors;
public class SortByValueExample {
    public static void main(String[] argv) {
        Map<String, Integer> unsortMap = new HashMap<>();
        unsortMap.put("z", 10);
        unsortMap.put("b", 5);
        unsortMap.put("a", 6);
        unsortMap.put("c", 20);
        unsortMap.put("d", 1);
        unsortMap.put("e", 7);
        unsortMap.put("y", 8);
        unsortMap.put("n", 99);
        unsortMap.put("g", 50);
        unsortMap.put("m", 2);
        unsortMap.put("f", 9);
        System.out.println("Original...");
        System.out.println(unsortMap);
        //sort by values, and reserve it, 10,9,8,7,6...
        Map<String, Integer> result = unsortMap.entrySet().stream()
                .sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,
                        (oldValue, newValue) -> oldValue, LinkedHashMap::new));
        //Alternative way
        Map<String, Integer> result2 = new LinkedHashMap<>();
        unsortMap.entrySet().stream()
                .sorted(Map.Entry.<String, Integer>comparingByValue().reversed())
                .forEachOrdered(x -> result2.put(x.getKey(), x.getValue()));
        System.out.println("Sorted...");
        System.out.println(result);
        System.out.println(result2);
    }
}

输出量

Original...
{a=6, b=5, c=20, d=1, e=7, f=9, g=50, y=8, z=10, m=2, n=99}
Sorted...
{n=99, g=50, c=20, z=10, f=9, y=8, e=7, a=6, b=5, m=2, d=1}
{n=99, g=50, c=20, z=10, f=9, y=8, e=7, a=6, b=5, m=2, d=1}

4. Map <对象，对象>

Stream无法直接对Map<Object,Object>排序。要解决该问题，请将其转换为Map<String,String> ，请查看以下示例。

注意
如果您有更好的想法进行排序，请告诉我。

DisplayApp.java

package com.mkyong;
import java.util.LinkedHashMap;
import java.util.Map;
import java.util.Properties;
import java.util.Set;
import java.util.stream.Collectors;
public class DisplayApp {
    public static void main(String[] args) {
        Properties properties = System.getProperties();
        // not easy to sort this
        Set<Map.Entry<Object, Object>> entries = properties.entrySet();
        LinkedHashMap<String, String> collect = entries.stream()
                //Map<String, String>
                .collect(Collectors.toMap(k -> (String) k.getKey(), e -> (String) e.getValue()))
                .entrySet()
                .stream()
                .sorted(Map.Entry.comparingByKey())
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,
                        (oldValue, newValue) -> oldValue, LinkedHashMap::new));
        collect.forEach((k, v) -> System.out.println(k + ":" + v));
    }
}