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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。
在java 8中按照此步骤对map进行排序.
- 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 将返回一个 HashMap。在这里通过使用toMap
方法的另一个变体来处理重复问题,它允许我们指定一个合并方法((oldValue, newValue) -> oldValue
)。这个合并方法允许用户他们指定想如何处理多个值关联到同一个键的冲突。在下面展示的代码中,我们只是使用了新的值,当然你也可以编写一个智能的算法来处理冲突。
- private static Map<String, Task> taskMap_duplicates(List<Task> tasks) {
- return tasks.stream().collect(toMap(Task::getTitle, identity(), (t1, t2) -> t2));
- }
你可以通过使用toMap
方法的第三个变体来指定其他的映射实现。这需要你指定将用来存储结果的Map和Supplier。
- public Map<String, Task> collectToMap(List<Task> tasks) {
- return tasks.stream().collect(toMap(Task::getTitle, identity(), (t1, t2) -> t2, LinkedHashMap::new));
- }
- 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);
- }
- }
- 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);
- }
- }
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