我对Java比较陌生,经常发现需要对值进行Map<Key,Value>排序。
由于这些值不是唯一的,我发现自己将keySet转换为一个数组,并使用自定义比较器通过数组排序对该数组进行排序,该比较器根据与该键关联的值进行排序。
有没有更简单的方法?
我对Java比较陌生,经常发现需要对值进行Map<Key,Value>排序。
由于这些值不是唯一的,我发现自己将keySet转换为一个数组,并使用自定义比较器通过数组排序对该数组进行排序,该比较器根据与该键关联的值进行排序。
有没有更简单的方法?
当前回答
public class Test {
public static void main(String[] args) {
TreeMap<Integer, String> hm=new TreeMap();
hm.put(3, "arun singh");
hm.put(5, "vinay singh");
hm.put(1, "bandagi singh");
hm.put(6, "vikram singh");
hm.put(2, "panipat singh");
hm.put(28, "jakarta singh");
ArrayList<String> al=new ArrayList(hm.values());
Collections.sort(al, new myComparator());
System.out.println("//sort by values \n");
for(String obj: al){
for(Map.Entry<Integer, String> map2:hm.entrySet()){
if(map2.getValue().equals(obj)){
System.out.println(map2.getKey()+" "+map2.getValue());
}
}
}
}
}
class myComparator implements Comparator{
@Override
public int compare(Object o1, Object o2) {
String o3=(String) o1;
String o4 =(String) o2;
return o3.compareTo(o4);
}
}
输出,输出=
//sort by values
3 arun singh
1 bandagi singh
28 jakarta singh
2 panipat singh
6 vikram singh
5 vinay singh
其他回答
主要问题。如果您使用第一个答案(Google将您带到这里),请更改比较器以添加等号子句,否则无法按键从sorted_map中获取值:
public int compare(String a, String b) {
if (base.get(a) > base.get(b)) {
return 1;
} else if (base.get(a) < base.get(b)){
return -1;
}
return 0;
// returning 0 would merge keys
}
HashMap<String,Long>的最简单的暴力排序HashMap方法:您可以复制粘贴它,然后这样使用:
public class Test {
public static void main(String[] args) {
HashMap<String, Long> hashMap = new HashMap<>();
hashMap.put("Cat", (long) 4);
hashMap.put("Human", (long) 2);
hashMap.put("Dog", (long) 4);
hashMap.put("Fish", (long) 0);
hashMap.put("Tree", (long) 1);
hashMap.put("Three-legged-human", (long) 3);
hashMap.put("Monkey", (long) 2);
System.out.println(hashMap); //{Human=2, Cat=4, Three-legged-human=3, Monkey=2, Fish=0, Tree=1, Dog=4}
System.out.println(sortHashMap(hashMap)); //{Cat=4, Dog=4, Three-legged-human=3, Human=2, Monkey=2, Tree=1, Fish=0}
}
public LinkedHashMap<String, Long> sortHashMap(HashMap<String, Long> unsortedMap) {
LinkedHashMap<String, Long> result = new LinkedHashMap<>();
//add String keys to an array: the array would get sorted, based on those keys' values
ArrayList<String> sortedKeys = new ArrayList<>();
for (String key: unsortedMap.keySet()) {
sortedKeys.add(key);
}
//sort the ArrayList<String> of keys
for (int i=0; i<unsortedMap.size(); i++) {
for (int j=1; j<sortedKeys.size(); j++) {
if (unsortedMap.get(sortedKeys.get(j)) > unsortedMap.get(sortedKeys.get(j-1))) {
String temp = sortedKeys.get(j);
sortedKeys.set(j, sortedKeys.get(j-1));
sortedKeys.set(j-1, temp);
}
}
}
// construct the result Map
for (String key: sortedKeys) {
result.put(key, unsortedMap.get(key));
}
return result;
}
}
使用java.util.TreeMap。
映射根据其键的自然顺序进行排序,或由创建映射时提供的Comparator进行排序,具体取决于使用的构造函数
使用LinkedList
//Create a list by HashMap
List<Map.Entry<String, Double>> list = new LinkedList<>(hashMap.entrySet());
//Sorting the list
Collections.sort(list, new Comparator<Map.Entry<String, Double>>() {
public int compare(Map.Entry<String, Double> o1, Map.Entry<String, Double> o2) {
return (o1.getValue()).compareTo(o2.getValue());
}
});
//put data from sorted list to hashmap
HashMap<String, Double> sortedData = new LinkedHashMap<>();
for (Map.Entry<String, Double> data : list) {
sortedData.put(data.getKey(), data.getValue());
}
System.out.print(sortedData);
在Java 8及以上版本中对任何地图进行排序的简单方法
Map<String, Object> mapToSort = new HashMap<>();
List<Map.Entry<String, Object>> list = new LinkedList<>(mapToSort.entrySet());
Collections.sort(list, Comparator.comparing(o -> o.getValue().getAttribute()));
HashMap<String, Object> sortedMap = new LinkedHashMap<>();
for (Map.Entry<String, Object> map : list) {
sortedMap.put(map.getKey(), map.getValue());
}
如果您使用的是Java 7及以下版本
Map<String, Object> mapToSort = new HashMap<>();
List<Map.Entry<String, Object>> list = new LinkedList<>(mapToSort.entrySet());
Collections.sort(list, new Comparator<Map.Entry<String, Object>>() {
@Override
public int compare(Map.Entry<String, Object> o1, Map.Entry<String, Object> o2) {
return o1.getValue().getAttribute().compareTo(o2.getValue().getAttribute());
}
});
HashMap<String, Object> sortedMap = new LinkedHashMap<>();
for (Map.Entry<String, Object> map : list) {
sortedMap.put(map.getKey(), map.getValue());
}