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
一些简单的更改,以便具有具有重复值的对的排序映射。在比较方法(类ValueComparator)中,当值相等时,不返回0,而是返回比较2个键的结果。关键点在地图中是不同的,因此您可以成功地保留重复的值(顺便按关键点排序)。因此,上面的示例可以这样修改:
public int compare(Object a, Object b) {
if((Double)base.get(a) < (Double)base.get(b)) {
return 1;
} else if((Double)base.get(a) == (Double)base.get(b)) {
return ((String)a).compareTo((String)b);
} else {
return -1;
}
}
}
其他回答
因为地图是无序的要对其进行排序,我们可以执行以下操作
Map<String, String> map= new TreeMap<String, String>(unsortMap);
您应该注意,与哈希映射不同,树映射保证其元素将按升序键排序。
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));
}
}
从…起http://www.programmersheaven.com/download/49349/download.aspx
private static <K, V> Map<K, V> sortByValue(Map<K, V> map) {
List<Entry<K, V>> list = new LinkedList<>(map.entrySet());
Collections.sort(list, new Comparator<Object>() {
@SuppressWarnings("unchecked")
public int compare(Object o1, Object o2) {
return ((Comparable<V>) ((Map.Entry<K, V>) (o1)).getValue()).compareTo(((Map.Entry<K, V>) (o2)).getValue());
}
});
Map<K, V> result = new LinkedHashMap<>();
for (Iterator<Entry<K, V>> it = list.iterator(); it.hasNext();) {
Map.Entry<K, V> entry = (Map.Entry<K, V>) it.next();
result.put(entry.getKey(), entry.getValue());
}
return result;
}
这太复杂了。地图不应该按价值排序。最简单的方法是创建自己的类,以满足您的需求。
在下面的示例中,您应该在*所在的位置添加TreeMap比较器。但通过javaAPI,它只提供比较器键,而不提供值。此处所述的所有示例均基于2个地图。一个哈希和一个新树。这很奇怪。
示例:
Map<Driver driver, Float time> map = new TreeMap<Driver driver, Float time>(*);
因此,通过以下方式将地图更改为集合:
ResultComparator rc = new ResultComparator();
Set<Results> set = new TreeSet<Results>(rc);
您将创建类Results,
public class Results {
private Driver driver;
private Float time;
public Results(Driver driver, Float time) {
this.driver = driver;
this.time = time;
}
public Float getTime() {
return time;
}
public void setTime(Float time) {
this.time = time;
}
public Driver getDriver() {
return driver;
}
public void setDriver (Driver driver) {
this.driver = driver;
}
}
以及Comparator类:
public class ResultsComparator implements Comparator<Results> {
public int compare(Results t, Results t1) {
if (t.getTime() < t1.getTime()) {
return 1;
} else if (t.getTime() == t1.getTime()) {
return 0;
} else {
return -1;
}
}
}
这样,您可以轻松添加更多依赖项。
最后一点,我将添加简单迭代器:
Iterator it = set.iterator();
while (it.hasNext()) {
Results r = (Results)it.next();
System.out.println( r.getDriver().toString
//or whatever that is related to Driver class -getName() getSurname()
+ " "
+ r.getTime()
);
}
当然,Stephen的解决方案真的很棒,但对于那些不会使用Guava的人来说:
这是我的解决方案,用于按值对地图进行排序。此解决方案处理两倍相同值等情况。。。
// If you want to sort a map by value, and if there can be twice the same value:
// here is your original map
Map<String,Integer> mapToSortByValue = new HashMap<String, Integer>();
mapToSortByValue.put("A", 3);
mapToSortByValue.put("B", 1);
mapToSortByValue.put("C", 3);
mapToSortByValue.put("D", 5);
mapToSortByValue.put("E", -1);
mapToSortByValue.put("F", 1000);
mapToSortByValue.put("G", 79);
mapToSortByValue.put("H", 15);
// Sort all the map entries by value
Set<Map.Entry<String,Integer>> set = new TreeSet<Map.Entry<String,Integer>>(
new Comparator<Map.Entry<String,Integer>>(){
@Override
public int compare(Map.Entry<String,Integer> obj1, Map.Entry<String,Integer> obj2) {
Integer val1 = obj1.getValue();
Integer val2 = obj2.getValue();
// DUPLICATE VALUE CASE
// If the values are equals, we can't return 0 because the 2 entries would be considered
// as equals and one of them would be deleted (because we use a set, no duplicate, remember!)
int compareValues = val1.compareTo(val2);
if ( compareValues == 0 ) {
String key1 = obj1.getKey();
String key2 = obj2.getKey();
int compareKeys = key1.compareTo(key2);
if ( compareKeys == 0 ) {
// what you return here will tell us if you keep REAL KEY-VALUE duplicates in your set
// if you want to, do whatever you want but do not return 0 (but don't break the comparator contract!)
return 0;
}
return compareKeys;
}
return compareValues;
}
}
);
set.addAll(mapToSortByValue.entrySet());
// OK NOW OUR SET IS SORTED COOL!!!!
// And there's nothing more to do: the entries are sorted by value!
for ( Map.Entry<String,Integer> entry : set ) {
System.out.println("Set entries: " + entry.getKey() + " -> " + entry.getValue());
}
// But if you add them to an hashmap
Map<String,Integer> myMap = new HashMap<String,Integer>();
// When iterating over the set the order is still good in the println...
for ( Map.Entry<String,Integer> entry : set ) {
System.out.println("Added to result map entries: " + entry.getKey() + " " + entry.getValue());
myMap.put(entry.getKey(), entry.getValue());
}
// But once they are in the hashmap, the order is not kept!
for ( Integer value : myMap.values() ) {
System.out.println("Result map values: " + value);
}
// Also this way doesn't work:
// Logic because the entryset is a hashset for hashmaps and not a treeset
// (and even if it was a treeset, it would be on the keys only)
for ( Map.Entry<String,Integer> entry : myMap.entrySet() ) {
System.out.println("Result map entries: " + entry.getKey() + " -> " + entry.getValue());
}
// CONCLUSION:
// If you want to iterate on a map ordered by value, you need to remember:
// 1) Maps are only sorted by keys, so you can't sort them directly by value
// 2) So you simply CAN'T return a map to a sortMapByValue function
// 3) You can't reverse the keys and the values because you have duplicate values
// This also means you can't neither use Guava/Commons bidirectionnal treemaps or stuff like that
// SOLUTIONS
// So you can:
// 1) only sort the values which is easy, but you loose the key/value link (since you have duplicate values)
// 2) sort the map entries, but don't forget to handle the duplicate value case (like i did)
// 3) if you really need to return a map, use a LinkedHashMap which keep the insertion order
执行官:http://www.ideone.com/dq3Lu
输出:
Set entries: E -> -1
Set entries: B -> 1
Set entries: A -> 3
Set entries: C -> 3
Set entries: D -> 5
Set entries: H -> 15
Set entries: G -> 79
Set entries: F -> 1000
Added to result map entries: E -1
Added to result map entries: B 1
Added to result map entries: A 3
Added to result map entries: C 3
Added to result map entries: D 5
Added to result map entries: H 15
Added to result map entries: G 79
Added to result map entries: F 1000
Result map values: 5
Result map values: -1
Result map values: 1000
Result map values: 79
Result map values: 3
Result map values: 1
Result map values: 3
Result map values: 15
Result map entries: D -> 5
Result map entries: E -> -1
Result map entries: F -> 1000
Result map entries: G -> 79
Result map entries: A -> 3
Result map entries: B -> 1
Result map entries: C -> 3
Result map entries: H -> 15
希望它能帮助一些人