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
使用通用比较器,例如:
final class MapValueComparator<K,V extends Comparable<V>> implements Comparator<K> {
private final Map<K,V> map;
private MapValueComparator() {
super();
}
public MapValueComparator(Map<K,V> map) {
this();
this.map = map;
}
public int compare(K o1, K o2) {
return map.get(o1).compareTo(map.get(o2));
}
}
其他回答
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;
}
}
我合并了user157196和Carter Page的解决方案:
class MapUtil {
public static <K, V extends Comparable<? super V>> Map<K, V> sortByValue( Map<K, V> map ){
ValueComparator<K,V> bvc = new ValueComparator<K,V>(map);
TreeMap<K,V> sorted_map = new TreeMap<K,V>(bvc);
sorted_map.putAll(map);
return sorted_map;
}
}
class ValueComparator<K, V extends Comparable<? super V>> implements Comparator<K> {
Map<K, V> base;
public ValueComparator(Map<K, V> base) {
this.base = base;
}
public int compare(K a, K b) {
int result = (base.get(a).compareTo(base.get(b)));
if (result == 0) result=1;
// returning 0 would merge keys
return result;
}
}
对于Java 8,您可以使用streams api以一种明显不那么冗长的方式来实现:
Map<K, V> sortedMap = map.entrySet().stream()
.sorted(Entry.comparingByValue())
.collect(Collectors.toMap(Entry::getKey, Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));
map = your hashmap;
List<Map.Entry<String, Integer>> list = new LinkedList<Map.Entry<String, Integer>>(map.entrySet());
Collections.sort(list, new cm());//IMP
HashMap<String, Integer> sorted = new LinkedHashMap<String, Integer>();
for(Map.Entry<String, Integer> en: list){
sorted.put(en.getKey(),en.getValue());
}
System.out.println(sorted);//sorted hashmap
创建新类
class cm implements Comparator<Map.Entry<String, Integer>>{
@Override
public int compare(Map.Entry<String, Integer> a,
Map.Entry<String, Integer> b)
{
return (a.getValue()).compareTo(b.getValue());
}
}
以下是Java8的代码,使用算盘
Map<String, Integer> map = N.asMap("a", 2, "b", 3, "c", 1, "d", 2);
Map<String, Integer> sortedMap = Stream.of(map.entrySet()).sorted(Map.Entry.comparingByValue()).toMap(e -> e.getKey(), e -> e.getValue(),
LinkedHashMap::new);
N.println(sortedMap);
// output: {c=1, a=2, d=2, b=3}
声明:我是算盘通用的开发者。