Java 8 Map 排序

Java 8的Stream例子Map，按键或按值排序。

快速说明

在Java 8中排序地图的步骤

1. 将map转换为流
2. 排序
3. Collect and return 新的LinkedHashMap

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

默认情况下Collectors.toMap会返回一个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);  }  }

Output

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

排序...
{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);  } }

Output

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

排序...
{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}