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
其他回答
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()
);
}
当你有两个相等的项目时，投票给最多的答案不起作用。TreeMap保留相等的值。
示例：未排序地图
key/value: D/67.3 key/value: A/99.5 key/value: B/67.4 key/value: C/67.5 key/value: E/99.5
后果
key/value: A/99.5 key/value: C/67.5 key/value: B/67.4 key/value: D/67.3
所以省略了E！！
对我来说，它可以很好地调整比较器，如果它等于，则不返回0，而是返回-1。
在示例中：
类ValueComparator实现Comparator{地图库；公共ValueComparator（地图库）{this.base=基数；}public int compare（对象a，对象b）{如果（（双）base.get（a）<（双）base.get（b））{返回1；}否则如果（（双）base.get（a）==（双）base.get（b））{返回-1；}其他{返回-1；}}}
现在它返回：
未排序地图：
key/value: D/67.3 key/value: A/99.5 key/value: B/67.4 key/value: C/67.5 key/value: E/99.5
结果：
key/value: A/99.5 key/value: E/99.5 key/value: C/67.5 key/value: B/67.4 key/value: D/67.3
作为对《外国人》的回应（2011年11月22日）：我将此解决方案用于整数Id和名称的映射，但想法是相同的，因此上面的代码可能不正确（我将在测试中编写并给您正确的代码），这是基于上面解决方案的map排序代码：
package nl.iamit.util;
import java.util.Comparator;
import java.util.Map;
public class Comparators {
public static class MapIntegerStringComparator implements Comparator {
Map<Integer, String> base;
public MapIntegerStringComparator(Map<Integer, String> base) {
this.base = base;
}
public int compare(Object a, Object b) {
int compare = ((String) base.get(a))
.compareTo((String) base.get(b));
if (compare == 0) {
return -1;
}
return compare;
}
}
}
这是测试类（我刚刚测试了它，这适用于Integer，StringMap：
package test.nl.iamit.util;
import java.util.HashMap;
import java.util.TreeMap;
import nl.iamit.util.Comparators;
import org.junit.Test;
import static org.junit.Assert.assertArrayEquals;
public class TestComparators {
@Test
public void testMapIntegerStringComparator(){
HashMap<Integer, String> unSoretedMap = new HashMap<Integer, String>();
Comparators.MapIntegerStringComparator bvc = new Comparators.MapIntegerStringComparator(
unSoretedMap);
TreeMap<Integer, String> sorted_map = new TreeMap<Integer, String>(bvc);
//the testdata:
unSoretedMap.put(new Integer(1), "E");
unSoretedMap.put(new Integer(2), "A");
unSoretedMap.put(new Integer(3), "E");
unSoretedMap.put(new Integer(4), "B");
unSoretedMap.put(new Integer(5), "F");
sorted_map.putAll(unSoretedMap);
Object[] targetKeys={new Integer(2),new Integer(4),new Integer(3),new Integer(1),new Integer(5) };
Object[] currecntKeys=sorted_map.keySet().toArray();
assertArrayEquals(targetKeys,currecntKeys);
}
}
以下是地图比较器的代码：
public static class MapStringDoubleComparator implements Comparator {
Map<String, Double> base;
public MapStringDoubleComparator(Map<String, Double> base) {
this.base = base;
}
//note if you want decending in stead of ascending, turn around 1 and -1
public int compare(Object a, Object b) {
if ((Double) base.get(a) == (Double) base.get(b)) {
return 0;
} else if((Double) base.get(a) < (Double) base.get(b)) {
return -1;
}else{
return 1;
}
}
}
这是一个测试用例：
@Test
public void testMapStringDoubleComparator(){
HashMap<String, Double> unSoretedMap = new HashMap<String, Double>();
Comparators.MapStringDoubleComparator bvc = new Comparators.MapStringDoubleComparator(
unSoretedMap);
TreeMap<String, Double> sorted_map = new TreeMap<String, Double>(bvc);
//the testdata:
unSoretedMap.put("D",new Double(67.3));
unSoretedMap.put("A",new Double(99.5));
unSoretedMap.put("B",new Double(67.4));
unSoretedMap.put("C",new Double(67.5));
unSoretedMap.put("E",new Double(99.5));
sorted_map.putAll(unSoretedMap);
Object[] targetKeys={"D","B","C","E","A"};
Object[] currecntKeys=sorted_map.keySet().toArray();
assertArrayEquals(targetKeys,currecntKeys);
}
当然，你可以让它更通用，但我只需要一个案例（地图）
由于TreeMap<>不适用于可以相等的值，因此我使用了以下方法：
private <K, V extends Comparable<? super V>> List<Entry<K, V>> sort(Map<K, V> map) {
List<Map.Entry<K, V>> list = new LinkedList<Map.Entry<K, V>>(map.entrySet());
Collections.sort(list, new Comparator<Map.Entry<K, V>>() {
public int compare(Map.Entry<K, V> o1, Map.Entry<K, V> o2) {
return o1.getValue().compareTo(o2.getValue());
}
});
return list;
}
您可能希望将列表放在LinkedHashMap中，但若您只打算立即对其进行迭代，那个么这是多余的。。。
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。
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
