JDK8 新特性-Map对key和value分别排序实现

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在Java 8 中使用Stream 例子对一个 Map 进行按照keys或者values排序.

1. 快速入门

在java 8中按照此步骤对map进行排序.

  1. 将 Map 转换为 Stream
  2. 对其进行排序
  3. Collect and return a new LinkedHashMap (保持顺序)
Map result = map.entrySet().stream()	.sorted(Map.Entry.comparingByKey())	.collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,	(oldValue, newValue) -> oldValue, LinkedHashMap::new));

P.S 默认情况下, Collectors.toMap 将返回一个 HashMap

2. 按照key排序

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:

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. 按照value排序​​​​​​​

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:

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}