Java8 对Map进行排序

版权声明：“假装、是个有灵魂的程序员” —— Go Big Or Go Home https://blog.csdn.net/u011663149/article/details/88558758

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前言：

    随着Java 8 的Stream之后推出后可以很优雅实现Map的排序。

示例：

final Map<String, Integer> wordCounts = new HashMap<>();
wordCounts.put("USA", 100);
wordCounts.put("jobs", 200);
wordCounts.put("software", 50);
wordCounts.put("technology", 70);
wordCounts.put("opportunity", 200);

//{USA=100, software=50, jobs=200, opportunity=200, technology=70}

按升序对值进行排序，使用LinkedHashMap存储排序结果来保留结果映射中元素的顺序

1、正向 
final Map<String, Integer> sortedByCount = wordCounts.entrySet()
                .stream()
                .sorted(Map.Entry.comparingByValue())
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));

//{software=50, technology=70, USA=100, jobs=200, opportunity=200}

2、反向 reversed
final Map<String, Integer> sortedByCount1 = wordCounts.entrySet()
        .stream()
        .sorted((Map.Entry.<String, Integer>comparingByValue().reversed()))
        .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));

//{jobs=200, opportunity=200, USA=100, technology=70, software=50}

推荐

sorted()方法将Comparator作为参数使用任何类型的值对映射进行排序。上面的排序可以用Comparator写成：

//正向 
  final Map<String, Integer> sortedByCount3 = wordCounts.entrySet()
        .stream()
        .sorted((e1, e2) -> e1.getValue().compareTo(e2.getValue()))
        .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));
//{software=50, technology=70, USA=100, jobs=200, opportunity=200}


//反向 == reversed()
final Map<String, Integer> sortedByCount2 = wordCounts.entrySet()
                .stream()
                .sorted((e1, e2) -> e2.getValue().compareTo(e1.getValue()))
                .collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));

//{jobs=200, opportunity=200, USA=100, technology=70, software=50}

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