按值</key,>对Map <key,value =“”>进行排序

时间:2008-09-20 21:03:23

标签: java sorting dictionary collections

我对Java比较陌生,经常发现我需要对值Map<Key, Value>进行排序。

由于值不是唯一的,我发现自己将keySet转换为array,并使用自定义比较器<数组排序对该数组进行排序< / strong>对与键相关联的值进行排序。

有更简单的方法吗?

60 个答案:

答案 0 :(得分:843)

这是一个通用友好版本:

public class MapUtil {
    public static <K, V extends Comparable<? super V>> Map<K, V> sortByValue(Map<K, V> map) {
        List<Entry<K, V>> list = new ArrayList<>(map.entrySet());
        list.sort(Entry.comparingByValue());

        Map<K, V> result = new LinkedHashMap<>();
        for (Entry<K, V> entry : list) {
            result.put(entry.getKey(), entry.getValue());
        }

        return result;
    }
}

答案 1 :(得分:408)

重要提示:

此代码可能会以多种方式中断。如果您打算使用提供的代码,请务必阅读注释以了解其含义。例如,不能再通过其密钥检索值。 (get始终返回null。)


似乎比上述所有内容都容易得多。使用TreeMap如下:

public class Testing {
    public static void main(String[] args) {
        HashMap<String, Double> map = new HashMap<String, Double>();
        ValueComparator bvc = new ValueComparator(map);
        TreeMap<String, Double> sorted_map = new TreeMap<String, Double>(bvc);

        map.put("A", 99.5);
        map.put("B", 67.4);
        map.put("C", 67.4);
        map.put("D", 67.3);

        System.out.println("unsorted map: " + map);
        sorted_map.putAll(map);
        System.out.println("results: " + sorted_map);
    }
}

class ValueComparator implements Comparator<String> {
    Map<String, Double> base;

    public ValueComparator(Map<String, Double> base) {
        this.base = base;
    }

    // Note: this comparator imposes orderings that are inconsistent with
    // equals.
    public int compare(String a, String b) {
        if (base.get(a) >= base.get(b)) {
            return -1;
        } else {
            return 1;
        } // returning 0 would merge keys
    }
}

输出:

unsorted map: {D=67.3, A=99.5, B=67.4, C=67.4}
results: {D=67.3, B=67.4, C=67.4, A=99.5}

答案 2 :(得分:283)

Java 8提供了一个新答案:将条目转换为流,并使用Map.Entry中的比较器组合器:

Stream<Map.Entry<K,V>> sorted =
    map.entrySet().stream()
       .sorted(Map.Entry.comparingByValue());

这将允许您使用按值的升序排序的条目。如果要降序值,只需反转比较器:

Stream<Map.Entry<K,V>> sorted =
    map.entrySet().stream()
       .sorted(Collections.reverseOrder(Map.Entry.comparingByValue()));

如果值不具有可比性,您可以传递一个明确的比较器:

Stream<Map.Entry<K,V>> sorted =
    map.entrySet().stream()
       .sorted(Map.Entry.comparingByValue(comparator));

然后,您可以继续使用其他流操作来使用数据。例如,如果您想要新地图中的前10名:

Map<K,V> topTen =
    map.entrySet().stream()
       .sorted(Map.Entry.comparingByValue(Comparator.reverseOrder()))
       .limit(10)
       .collect(Collectors.toMap(
          Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));

或打印到System.out

map.entrySet().stream()
   .sorted(Map.Entry.comparingByValue())
   .forEach(System.out::println);

答案 3 :(得分:209)

三个1行答案......

我会使用 Google Collections Guava来执行此操作 - 如果您的值为Comparable,则可以使用

valueComparator = Ordering.natural().onResultOf(Functions.forMap(map))

这将为地图创建一个函数(对象)[将任意键作为输入,返回相应的值],然后对它们[值]应用自然(可比较)排序。

如果它们不具有可比性,那么你需要按照

的方式做一些事情
valueComparator = Ordering.from(comparator).onResultOf(Functions.forMap(map)) 

这些可以应用于TreeMap(Ordering extends Comparator)或LinkedHashMap after some sorting

NB :如果您打算使用TreeMap,请记住,如果比较== 0,那么该项目已经在列表中(如果您有多个值可以比较相同)。为了缓解这个问题,你可以像往常一样将你的密钥添加到比较器中(假设你的密钥和值是Comparable):

valueComparator = Ordering.natural().onResultOf(Functions.forMap(map)).compound(Ordering.natural())

= 将自然排序应用于键映射的值,并使用键的自然顺序复合

请注意,如果您的密钥与0比较,这仍然不起作用,但这对于大多数comparable项来说已足够(因为hashCodeequalscompareTo是经常同步...)

请参阅Ordering.onResultOf()Functions.forMap()

实施

现在我们已经有了一个可以满足我们想要的比较器,我们需要从中获得结果。

map = ImmutableSortedMap.copyOf(myOriginalMap, valueComparator);

现在这很可能会起作用,但是:

    如果给出完整的完成地图,
  1. 需要完成
  2. 不要在TreeMap上尝试上面的比较器;
  3. 之后没有必要尝试比较插入的密钥,直到它之后没有值。 对于我来说,第1点对我来说是一个破坏性的事情; google集合非常懒惰(这很好:你可以在瞬间完成所有操作;真正的工作是在你开始使用结果时完成的),这需要复制整个地图! / p>

    “完整”答案/按值分类的实时分类地图

    不过不用担心;如果你对以这种方式排序的“实时”地图足够痴迷,你可以解决上述问题中的一个而不是两个(!),如下所示:

    注意:这在2012年6月发生了重大变化 - 之前的代码永远不会起作用:需要内部HashMap来查找值,而不会在TreeMap.get() - &gt;之间创建无限循环。 compare()compare() - &gt; get()

    import static org.junit.Assert.assertEquals;
    
    import java.util.HashMap;
    import java.util.Map;
    import java.util.TreeMap;
    
    import com.google.common.base.Functions;
    import com.google.common.collect.Ordering;
    
    class ValueComparableMap<K extends Comparable<K>,V> extends TreeMap<K,V> {
        //A map for doing lookups on the keys for comparison so we don't get infinite loops
        private final Map<K, V> valueMap;
    
        ValueComparableMap(final Ordering<? super V> partialValueOrdering) {
            this(partialValueOrdering, new HashMap<K,V>());
        }
    
        private ValueComparableMap(Ordering<? super V> partialValueOrdering,
                HashMap<K, V> valueMap) {
            super(partialValueOrdering //Apply the value ordering
                    .onResultOf(Functions.forMap(valueMap)) //On the result of getting the value for the key from the map
                    .compound(Ordering.natural())); //as well as ensuring that the keys don't get clobbered
            this.valueMap = valueMap;
        }
    
        public V put(K k, V v) {
            if (valueMap.containsKey(k)){
                //remove the key in the sorted set before adding the key again
                remove(k);
            }
            valueMap.put(k,v); //To get "real" unsorted values for the comparator
            return super.put(k, v); //Put it in value order
        }
    
        public static void main(String[] args){
            TreeMap<String, Integer> map = new ValueComparableMap<String, Integer>(Ordering.natural());
            map.put("a", 5);
            map.put("b", 1);
            map.put("c", 3);
            assertEquals("b",map.firstKey());
            assertEquals("a",map.lastKey());
            map.put("d",0);
            assertEquals("d",map.firstKey());
            //ensure it's still a map (by overwriting a key, but with a new value) 
            map.put("d", 2);
            assertEquals("b", map.firstKey());
            //Ensure multiple values do not clobber keys
            map.put("e", 2);
            assertEquals(5, map.size());
            assertEquals(2, (int) map.get("e"));
            assertEquals(2, (int) map.get("d"));
        }
     }
    

    当我们放置时,我们确保哈希映射具有比较器的值,然后放入TreeSet进行排序。但在此之前,我们检查哈希映射以查看密钥实际上并不重复。此外,我们创建的比较器还将包含密钥,以便重复值不会删除非重复键(由于==比较)。 这2项至关重要用于确保保留地图合同;如果你认为你不想那样,那么你几乎就是完全颠倒地图(到Map<V,K>)。

    构造函数需要被调用为

     new ValueComparableMap(Ordering.natural());
     //or
     new ValueComparableMap(Ordering.from(comparator));
    

答案 4 :(得分:182)

来自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;
}

答案 5 :(得分:48)

使用Java 8,您可以使用streams api以非常简洁的方式执行此操作:

Map<K, V> sortedMap = map.entrySet().stream()
                         .sorted(Entry.comparingByValue())
                         .collect(Collectors.toMap(Entry::getKey, Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));

答案 6 :(得分:31)

对键进行排序需要比较器查找每个比较的每个值。一个更具可扩展性的解决方案将直接使用entrySet,因为这样的值可以立即用于每次比较(尽管我没有用数字来支持)。

以下是此类事物的通用版本:

public static <K, V extends Comparable<? super V>> List<K> getKeysSortedByValue(Map<K, V> map) {
    final int size = map.size();
    final List<Map.Entry<K, V>> list = new ArrayList<Map.Entry<K, V>>(size);
    list.addAll(map.entrySet());
    final ValueComparator<V> cmp = new ValueComparator<V>();
    Collections.sort(list, cmp);
    final List<K> keys = new ArrayList<K>(size);
    for (int i = 0; i < size; i++) {
        keys.set(i, list.get(i).getKey());
    }
    return keys;
}

private static final class ValueComparator<V extends Comparable<? super V>>
                                     implements Comparator<Map.Entry<?, V>> {
    public int compare(Map.Entry<?, V> o1, Map.Entry<?, V> o2) {
        return o1.getValue().compareTo(o2.getValue());
    }
}

有很多方法可以减少上述解决方案的内存轮换。例如,创建的第一个ArrayList可以重新用作返回值;这需要抑制一些泛型警告,但对于可重用的库代码可能是值得的。此外,不必在每次调用时重新分配比较器。

这是一个效率更高但不太吸引人的版本:

public static <K, V extends Comparable<? super V>> List<K> getKeysSortedByValue2(Map<K, V> map) {
    final int size = map.size();
    final List reusedList = new ArrayList(size);
    final List<Map.Entry<K, V>> meView = reusedList;
    meView.addAll(map.entrySet());
    Collections.sort(meView, SINGLE);
    final List<K> keyView = reusedList;
    for (int i = 0; i < size; i++) {
        keyView.set(i, meView.get(i).getKey());
    }
    return keyView;
}

private static final Comparator SINGLE = new ValueComparator();

最后,如果您需要不断访问已排序的信息(而不是仅仅偶尔排序一次),则可以使用其他多地图。如果您需要更多详细信息,请与我们联系......

答案 7 :(得分:26)

commons-collections库包含一个名为TreeBidiMap的解决方案。或者,您可以查看Google Collections API。它有TreeMultimap你可以使用。

如果您不想使用这些框架......它们带有源代码。

答案 8 :(得分:25)

我查看了给定的答案,但是当很多键具有相同的值时,其中很多都比需要的更复杂或删除了地图元素。

这是一个我认为更合适的解决方案:

public static <K, V extends Comparable<V>> Map<K, V> sortByValues(final Map<K, V> map) {
    Comparator<K> valueComparator =  new Comparator<K>() {
        public int compare(K k1, K k2) {
            int compare = map.get(k2).compareTo(map.get(k1));
            if (compare == 0) return 1;
            else return compare;
        }
    };
    Map<K, V> sortedByValues = new TreeMap<K, V>(valueComparator);
    sortedByValues.putAll(map);
    return sortedByValues;
}

请注意,地图从最高值到最低值排序。

答案 9 :(得分:17)

使用Java 8中的新功能实现此目的:

import static java.util.Map.Entry.comparingByValue;
import static java.util.stream.Collectors.toList;

<K, V> List<Entry<K, V>> sort(Map<K, V> map, Comparator<? super V> comparator) {
    return map.entrySet().stream().sorted(comparingByValue(comparator)).collect(toList());
}

使用给定的比较器按条目对条目进行排序。或者,如果您的值可以相互比较,则不需要明确的比较器:

<K, V extends Comparable<? super V>> List<Entry<K, V>> sort(Map<K, V> map) {
    return map.entrySet().stream().sorted(comparingByValue()).collect(toList());
}

返回的列表是调用此方法时给定地图的快照,因此两者都不会反映对另一个的后续更改。对于地图的实时可迭代视图:

<K, V extends Comparable<? super V>> Iterable<Entry<K, V>> sort(Map<K, V> map) {
    return () -> map.entrySet().stream().sorted(comparingByValue()).iterator();
}

返回的iterable每次迭代时都会创建给定地图的新快照,因此除非进行并发修改,否则它将始终反映地图的当前状态。

答案 10 :(得分:15)

创建自定义比较器并在创建新TreeMap对象时使用它。

class MyComparator implements Comparator<Object> {

    Map<String, Integer> map;

    public MyComparator(Map<String, Integer> map) {
        this.map = map;
    }

    public int compare(Object o1, Object o2) {

        if (map.get(o2) == map.get(o1))
            return 1;
        else
            return ((Integer) map.get(o2)).compareTo((Integer)     
                                                            map.get(o1));

    }
}

在主要功能

中使用以下代码
    Map<String, Integer> lMap = new HashMap<String, Integer>();
    lMap.put("A", 35);
    lMap.put("B", 75);
    lMap.put("C", 50);
    lMap.put("D", 50);

    MyComparator comparator = new MyComparator(lMap);

    Map<String, Integer> newMap = new TreeMap<String, Integer>(comparator);
    newMap.putAll(lMap);
    System.out.println(newMap);

输出:

{B=75, D=50, C=50, A=35}

答案 11 :(得分:14)

虽然我同意对地图进行排序的不断需要可能是一种气味,但我认为以下代码是最简单的方法,而不使用不同的数据结构。

public class MapUtilities {

public static <K, V extends Comparable<V>> List<Entry<K, V>> sortByValue(Map<K, V> map) {
    List<Entry<K, V>> entries = new ArrayList<Entry<K, V>>(map.entrySet());
    Collections.sort(entries, new ByValue<K, V>());
    return entries;
}

private static class ByValue<K, V extends Comparable<V>> implements Comparator<Entry<K, V>> {
    public int compare(Entry<K, V> o1, Entry<K, V> o2) {
        return o1.getValue().compareTo(o2.getValue());
    }
}

}

这是一个令人尴尬的不完整的单元测试:

public class MapUtilitiesTest extends TestCase {
public void testSorting() {
    HashMap<String, Integer> map = new HashMap<String, Integer>();
    map.put("One", 1);
    map.put("Two", 2);
    map.put("Three", 3);

    List<Map.Entry<String, Integer>> sorted = MapUtilities.sortByValue(map);
    assertEquals("First", "One", sorted.get(0).getKey());
    assertEquals("Second", "Two", sorted.get(1).getKey());
    assertEquals("Third", "Three", sorted.get(2).getKey());
}

}

结果是Map.Entry对象的排序列表,您可以从中获取键和值。

答案 12 :(得分:11)

当你有两个等于的项目时,最多投票的答案不起作用。 TreeMap保留相同的值。

exmaple: 未分类的地图

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: C/67.5
key/value: B/67.4
key/value: D/67.3

离开E !!

对我来说,调整比较器是正常的,如果它等于不返回0而是-1。

示例中的

  
    

类ValueComparator实现Comparator {

         

地图基地;       public ValueComparator(Map base){           this.base = base;       }

         

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 -1;
} else {
  return -1;
}
         

}     }

  

现在它返回:

未分类的地图:

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

作为对外星人的回应(2011年11月22日): 我使用这个解决方案来获取I​​nteger Id和名称的映射,但是这个想法是一样的,所以上面的代码可能不正确(我会在测试中编写并给你正确的代码),这就是代码对于地图排序,基于上面的解决方案:

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(我刚测试过它,这适用于Integer,String Map:

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);
    }
}

这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);
}

of cource你可以使它更通用,但我只需要1个案例(地图)

答案 13 :(得分:11)

使用通用比较器,例如:

final class MapValueComparator<K,V extends Comparable<V>> implements Comparator<K> {

    private 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));
    }
}

答案 14 :(得分:9)

我建议使用Collections.sort,而不是使用Arrays.sort。实际上Collections.sort的作用是这样的:

public static <T extends Comparable<? super T>> void sort(List<T> list) {
    Object[] a = list.toArray();
    Arrays.sort(a);
    ListIterator<T> i = list.listIterator();
    for (int j=0; j<a.length; j++) {
        i.next();
        i.set((T)a[j]);
    }
}

只需在列表中调用toArray,然后使用Arrays.sort。这样,所有映射条目都将被复制三次:一次从映射到临时列表(无论是LinkedList还是ArrayList),然后到临时数组,最后到新映射。

我的解决方案省略了这一步,因为它不会创建不必要的LinkedList。以下是代码,通用友好和性能最佳:

public static <K, V extends Comparable<? super V>> Map<K, V> sortByValue(Map<K, V> map) 
{
    @SuppressWarnings("unchecked")
    Map.Entry<K,V>[] array = map.entrySet().toArray(new Map.Entry[map.size()]);

    Arrays.sort(array, new Comparator<Map.Entry<K, V>>() 
    {
        public int compare(Map.Entry<K, V> e1, Map.Entry<K, V> e2) 
        {
            return e1.getValue().compareTo(e2.getValue());
        }
    });

    Map<K, V> result = new LinkedHashMap<K, V>();
    for (Map.Entry<K, V> entry : array)
        result.put(entry.getKey(), entry.getValue());

    return result;
}

答案 15 :(得分:8)

这是Anthony的答案的变体,如果有重复的值,则无效:

public static <K, V extends Comparable<V>> Map<K, V> sortMapByValues(final Map<K, V> map) {
    Comparator<K> valueComparator =  new Comparator<K>() {
        public int compare(K k1, K k2) {
            final V v1 = map.get(k1);
            final V v2 = map.get(k2);

            /* Not sure how to handle nulls ... */
            if (v1 == null) {
                return (v2 == null) ? 0 : 1;
            }

            int compare = v2.compareTo(v1);
            if (compare != 0)
            {
                return compare;
            }
            else
            {
                Integer h1 = k1.hashCode();
                Integer h2 = k2.hashCode();
                return h2.compareTo(h1);
            }
        }
    };
    Map<K, V> sortedByValues = new TreeMap<K, V>(valueComparator);
    sortedByValues.putAll(map);
    return sortedByValues;
}

请注意,它是如何处理空值的。

这种方法的一个重要优点是它实际上返回了一个Map,与此处提供的其他一些解决方案不同。

答案 16 :(得分:7)

最佳方法

import java.util.ArrayList;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Set;
import java.util.Map.Entry; 

public class OrderByValue {

  public static void main(String a[]){
    Map<String, Integer> map = new HashMap<String, Integer>();
    map.put("java", 20);
    map.put("C++", 45);
    map.put("Unix", 67);
    map.put("MAC", 26);
    map.put("Why this kolavari", 93);
    Set<Entry<String, Integer>> set = map.entrySet();
    List<Entry<String, Integer>> list = new ArrayList<Entry<String, Integer>>(set);
    Collections.sort( list, new Comparator<Map.Entry<String, Integer>>()
    {
        public int compare( Map.Entry<String, Integer> o1, Map.Entry<String, Integer> o2 )
        {
            return (o1.getValue()).compareTo( o2.getValue() );//Ascending order
            //return (o2.getValue()).compareTo( o1.getValue() );//Descending order
        }
    } );
    for(Map.Entry<String, Integer> entry:list){
        System.out.println(entry.getKey()+" ==== "+entry.getValue());
    }
  }}

<强>输出

java ==== 20

MAC ==== 26

C++ ==== 45

Unix ==== 67

Why this kolavari ==== 93

答案 17 :(得分:7)

主要问题。如果您使用第一个答案(Google将您带到此处),请更改比较器以添加相等的子句,否则您无法通过键从sorted_map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
    }

答案 18 :(得分:6)

这个问题已经有很多答案了,但没有一个提供我正在寻找的东西,一个地图实现返回按关联值排序的键和条目,并将此属性维护为键,并在地图。两个other questions具体要求这一点。

我编写了一个解决此用例的通用友好示例。此实现不遵守Map接口的所有合同,例如反映从原始对象中的keySet()和entrySet()返回的集合中的值更改和删除。我觉得这样的解决方案太大而无法包含在Stack Overflow答案中。如果我设法创建一个更完整的实现,也许我会将它发布到Github,然后链接到这个答案的更新版本。

import java.util.*;

/**
 * A map where {@link #keySet()} and {@link #entrySet()} return sets ordered
 * by associated values based on the the comparator provided at construction
 * time. The order of two or more keys with identical values is not defined.
 * <p>
 * Several contracts of the Map interface are not satisfied by this minimal
 * implementation.
 */
public class ValueSortedMap<K, V> extends HashMap<K, V> {
    protected Map<V, Collection<K>> valueToKeysMap;

    // uses natural order of value object, if any
    public ValueSortedMap() {
        this((Comparator<? super V>) null);
    }

    public ValueSortedMap(Comparator<? super V> valueComparator) {
        this.valueToKeysMap = new TreeMap<V, Collection<K>>(valueComparator);
    }

    public boolean containsValue(Object o) {
        return valueToKeysMap.containsKey(o);
    }

    public V put(K k, V v) {
        V oldV = null;
        if (containsKey(k)) {
            oldV = get(k);
            valueToKeysMap.get(oldV).remove(k);
        }
        super.put(k, v);
        if (!valueToKeysMap.containsKey(v)) {
            Collection<K> keys = new ArrayList<K>();
            keys.add(k);
            valueToKeysMap.put(v, keys);
        } else {
            valueToKeysMap.get(v).add(k);
        }
        return oldV;
    }

    public void putAll(Map<? extends K, ? extends V> m) {
        for (Map.Entry<? extends K, ? extends V> e : m.entrySet())
            put(e.getKey(), e.getValue());
    }

    public V remove(Object k) {
        V oldV = null;
        if (containsKey(k)) {
            oldV = get(k);
            super.remove(k);
            valueToKeysMap.get(oldV).remove(k);
        }
        return oldV;
    }

    public void clear() {
        super.clear();
        valueToKeysMap.clear();
    }

    public Set<K> keySet() {
        LinkedHashSet<K> ret = new LinkedHashSet<K>(size());
        for (V v : valueToKeysMap.keySet()) {
            Collection<K> keys = valueToKeysMap.get(v);
            ret.addAll(keys);
        }
        return ret;
    }

    public Set<Map.Entry<K, V>> entrySet() {
        LinkedHashSet<Map.Entry<K, V>> ret = new LinkedHashSet<Map.Entry<K, V>>(size());
        for (Collection<K> keys : valueToKeysMap.values()) {
            for (final K k : keys) {
                final V v = get(k);
                ret.add(new Map.Entry<K,V>() {
                    public K getKey() {
                        return k;
                    }

                    public V getValue() {
                        return v;
                    }

                    public V setValue(V v) {
                        throw new UnsupportedOperationException();
                    }
                });
            }
        }
        return ret;
    }
}

答案 19 :(得分:5)

根据上下文,使用java.util.LinkedHashMap<T>来记住项目放入地图的顺序。否则,如果您需要根据其自然顺序对值进行排序,我建议您维护一个单独的列表,该列表可以通过Collections.sort()进行排序。

答案 20 :(得分:5)

由于 TreeMap&lt;&gt;对于可以相等的值,它不起作用 ,我使用了这个:

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,但如果您只是要迭代马上,这是多余的......

答案 21 :(得分:5)

这太复杂了。地图不应该按照Value对它们进行排序。最简单的方法是创建自己的类,以满足您的要求。

在示例中,您应该在*所在的位置添加TreeMap比较器。但是通过java API,它只为比较器提供键,而不是值。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>(*);

所以用这种方式将地图改成一组:

ResultComparator rc = new ResultComparator();
Set<Results> set = new TreeSet<Results>(rc);

您将创建课程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;
    }
}

和比较器类:

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;
        }
    }
}

通过这种方式,您可以轻松添加更多依赖项。

作为最后一点,我将添加简单的迭代器:

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()
        );
}

答案 22 :(得分:4)

Afaik最清洁的方式是使用集合来对值进行排序:

Map<String, Long> map = new HashMap<String, Long>();
// populate with data to sort on Value
// use datastructure designed for sorting

Queue queue = new PriorityQueue( map.size(), new MapComparable() );
queue.addAll( map.entrySet() );

// get a sorted map
LinkedHashMap<String, Long> linkedMap = new LinkedHashMap<String, Long>();

for (Map.Entry<String, Long> entry; (entry = queue.poll())!=null;) {
    linkedMap.put(entry.getKey(), entry.getValue());
}

public static class MapComparable implements Comparator<Map.Entry<String, Long>>{

  public int compare(Entry<String, Long> e1, Entry<String, Long> e2) {
    return e1.getValue().compareTo(e2.getValue());
  }
}

答案 23 :(得分:4)

一些简单的更改,以便使用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但返回比较2个键的结果。键在地图中是不同的,因此您可以成功保留重复值(顺便按键排序)。所以上面的例子可以修改如下:

    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;
        }
      }
    }

答案 24 :(得分:4)

肯定斯蒂芬的解决方案真的很棒,但对于那些不能使用番石榴的人来说:

这是我按地图分类的解决方案。 这个解决方案可以处理两倍相同值的情况......

// 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

希望它会帮助一些人

答案 25 :(得分:4)

这是一个OO解决方案(即,不使用static方法):

import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.LinkedHashMap;
import java.util.List;
import java.util.Map;

public class SortableValueMap<K, V extends Comparable<V>>
  extends LinkedHashMap<K, V> {
  public SortableValueMap() { }

  public SortableValueMap( Map<K, V> map ) {
    super( map );
  }

  public void sortByValue() {
    List<Map.Entry<K, V>> list = new LinkedList<Map.Entry<K, V>>( entrySet() );

    Collections.sort( list, new Comparator<Map.Entry<K, V>>() {
      public int compare( Map.Entry<K, V> entry1, Map.Entry<K, V> entry2 ) {
        return entry1.getValue().compareTo( entry2.getValue() );
      }
    });

    clear();

    for( Map.Entry<K, V> entry : list ) {
      put( entry.getKey(), entry.getValue() );
    }
  }

  private static void print( String text, Map<String, Double> map ) {
    System.out.println( text );

    for( String key : map.keySet() ) {
      System.out.println( "key/value: " + key + "/" + map.get( key ) );
    }
  }

  public static void main( String[] args ) {
    SortableValueMap<String, Double> map =
      new SortableValueMap<String, Double>();

    map.put( "A", 67.5 );
    map.put( "B", 99.5 );
    map.put( "C", 82.4 );
    map.put( "D", 42.0 );

    print( "Unsorted map", map );
    map.sortByValue();
    print( "Sorted map", map );
  }
}

特此捐赠给公共领域。

答案 26 :(得分:4)

基于@devinmoore代码,使用泛型并支持升序和降序的地图排序方法。

/**
 * Sort a map by it's keys in ascending order. 
 *  
 * @return new instance of {@link LinkedHashMap} contained sorted entries of supplied map.
 * @author Maxim Veksler
 */
public static <K, V> LinkedHashMap<K, V> sortMapByKey(final Map<K, V> map) {
    return sortMapByKey(map, SortingOrder.ASCENDING);
}

/**
 * Sort a map by it's values in ascending order.
 *  
 * @return new instance of {@link LinkedHashMap} contained sorted entries of supplied map.
 * @author Maxim Veksler
 */
public static <K, V> LinkedHashMap<K, V> sortMapByValue(final Map<K, V> map) {
    return sortMapByValue(map, SortingOrder.ASCENDING);
}

/**
 * Sort a map by it's keys.
 *  
 * @param sortingOrder {@link SortingOrder} enum specifying requested sorting order. 
 * @return new instance of {@link LinkedHashMap} contained sorted entries of supplied map.
 * @author Maxim Veksler
 */
public static <K, V> LinkedHashMap<K, V> sortMapByKey(final Map<K, V> map, final SortingOrder sortingOrder) {
    Comparator<Map.Entry<K, V>> comparator = new Comparator<Entry<K,V>>() {
        public int compare(Entry<K, V> o1, Entry<K, V> o2) {
            return comparableCompare(o1.getKey(), o2.getKey(), sortingOrder);
        }
    };

    return sortMap(map, comparator);
}

/**
 * Sort a map by it's values.
 *  
 * @param sortingOrder {@link SortingOrder} enum specifying requested sorting order. 
 * @return new instance of {@link LinkedHashMap} contained sorted entries of supplied map.
 * @author Maxim Veksler
 */
public static <K, V> LinkedHashMap<K, V> sortMapByValue(final Map<K, V> map, final SortingOrder sortingOrder) {
    Comparator<Map.Entry<K, V>> comparator = new Comparator<Entry<K,V>>() {
        public int compare(Entry<K, V> o1, Entry<K, V> o2) {
            return comparableCompare(o1.getValue(), o2.getValue(), sortingOrder);
        }
    };

    return sortMap(map, comparator);
}

@SuppressWarnings("unchecked")
private static <T> int comparableCompare(T o1, T o2, SortingOrder sortingOrder) {
    int compare = ((Comparable<T>)o1).compareTo(o2);

    switch (sortingOrder) {
    case ASCENDING:
        return compare;
    case DESCENDING:
        return (-1) * compare;
    }

    return 0;
}

/**
 * Sort a map by supplied comparator logic.
 *  
 * @return new instance of {@link LinkedHashMap} contained sorted entries of supplied map.
 * @author Maxim Veksler
 */
public static <K, V> LinkedHashMap<K, V> sortMap(final Map<K, V> map, final Comparator<Map.Entry<K, V>> comparator) {
    // Convert the map into a list of key,value pairs.
    List<Map.Entry<K, V>> mapEntries = new LinkedList<Map.Entry<K, V>>(map.entrySet());

    // Sort the converted list according to supplied comparator.
    Collections.sort(mapEntries, comparator);

    // Build a new ordered map, containing the same entries as the old map.  
    LinkedHashMap<K, V> result = new LinkedHashMap<K, V>(map.size() + (map.size() / 20));
    for(Map.Entry<K, V> entry : mapEntries) {
        // We iterate on the mapEntries list which is sorted by the comparator putting new entries into 
        // the targeted result which is a sorted map. 
        result.put(entry.getKey(), entry.getValue());
    }

    return result;
}

/**
 * Sorting order enum, specifying request result sort behavior.
 * @author Maxim Veksler
 *
 */
public static enum SortingOrder {
    /**
     * Resulting sort will be from smaller to biggest.
     */
    ASCENDING,
    /**
     * Resulting sort will be from biggest to smallest.
     */
    DESCENDING
}

答案 27 :(得分:4)

迟到。

随着Java-8的出现,我们可以以非常简单/简洁的方式使用流来进行数据操作。您可以使用流按值对地图条目进行排序,并创建LinkedHashMap,以保留插入顺序迭代。

例如:

LinkedHashMap sortedByValueMap = map.entrySet().stream()
                .sorted(comparing(Entry<Key,Value>::getValue).thenComparing(Entry::getKey))     //first sorting by Value, then sorting by Key(entries with same value)
                .collect(LinkedHashMap::new,(map,entry) -> map.put(entry.getKey(),entry.getValue()),LinkedHashMap::putAll);

对于反向排序,请替换:

comparing(Entry<Key,Value>::getValue).thenComparing(Entry::getKey)

comparing(Entry<Key,Value>::getValue).thenComparing(Entry::getKey).reversed()

答案 28 :(得分:3)

以下是Java 8的代码AbacusUtil

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AbacusUtil的开发者。

答案 29 :(得分:3)

如果您有重复的密钥且只有一小组数据(&lt; 1000)且您的代码对性能不重要,那么您可以执行以下操作:

Map<String,Integer> tempMap=new HashMap<String,Integer>(inputUnsortedMap);
LinkedHashMap<String,Integer> sortedOutputMap=new LinkedHashMap<String,Integer>();

for(int i=0;i<inputUnsortedMap.size();i++){
    Map.Entry<String,Integer> maxEntry=null;
    Integer maxValue=-1;
    for(Map.Entry<String,Integer> entry:tempMap.entrySet()){
        if(entry.getValue()>maxValue){
            maxValue=entry.getValue();
            maxEntry=entry;
        }
    }
    tempMap.remove(maxEntry.getKey());
    sortedOutputMap.put(maxEntry.getKey(),maxEntry.getValue());
}

inputUnsortedMap 是代码的输入。

变量 sortedOutputMap 将在迭代时按降序包含数据。要更改订单,只需更改&gt;到&lt;在if语句中。

不是最快的排序,但没有任何额外的依赖关系。

答案 30 :(得分:3)

使用 java 8 可以很容易地实现这一点

        @Bean
        @Qualifier("accountApiAuthenticationFilter")
        public TokenAuthenticationFilter accountApiAuthenticationFilter(TokenAuthenticationFailureHandler failureHandler,
                TokenAuthenticationSuccessHandler successHandler) throws Exception {
            TokenAuthenticationFilter filter = new TokenAuthenticationFilter();
            filter.setAuthenticationManager(authenticationManagerBean());
            filter.setAuthenticationFailureHandler(failureHandler);
            filter.setAuthenticationSuccessHandler(successHandler);
            filter.setAllowedTokenTypes(TokenType.APPLICATION, TokenType.BEARER);
            return filter;
        }
        
        @Override
        protected void configure(HttpSecurity http) throws Exception {
            http.antMatcher("/api/account/**")
                .addFilterBefore(accountApiAuthenticationFilter(null, null), UsernamePasswordAuthenticationFilter.class)
                .authorizeRequests()
                .antMatchers("**").hasAnyAuthority("READ_ALL", "BEARER_ACCESS")
                .anyRequest().authenticated()
                .and()
                .authenticationProvider(this.appProvider)
                .authenticationProvider(this.bearerProvider)
                .sessionManagement().sessionCreationPolicy(SessionCreationPolicy.STATELESS)
                .and()
                .exceptionHandling()
                .authenticationEntryPoint(new TokenAuthenticationEntryPoint())
                .and()
                .cors().disable()
                .formLogin().disable()
                .csrf().disable()
                .logout().disable();
        }

答案 31 :(得分:3)

您可以尝试Guava的多重映射:

TreeMap<Integer, Collection<String>> sortedMap = new TreeMap<>(
        Multimaps.invertFrom(Multimaps.forMap(originalMap), 
        ArrayListMultimap.<Integer, String>create()).asMap());

因此,您可以获得从原始值到与其对应的键集合的映射。即使有相同值的多个键,也可以使用此方法。

答案 32 :(得分:3)

我已经合并了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;
    }
}

答案 33 :(得分:2)

此方法仅用于此目的。 ('setback'是值必须实现java.util.Comparable接口

  /**

 * Sort a map according to values.

 * @param <K> the key of the map.
 * @param <V> the value to sort according to.
 * @param mapToSort the map to sort.

 * @return a map sorted on the values.

 */ 
public static <K, V extends Comparable< ? super V>> Map<K, V>
sortMapByValues(final Map <K, V> mapToSort)
{
    List<Map.Entry<K, V>> entries =
        new ArrayList<Map.Entry<K, V>>(mapToSort.size());  

    entries.addAll(mapToSort.entrySet());

    Collections.sort(entries,
                     new Comparator<Map.Entry<K, V>>()
    {
        @Override
        public int compare(
               final Map.Entry<K, V> entry1,
               final Map.Entry<K, V> entry2)
        {
            return entry1.getValue().compareTo(entry2.getValue());
        }
    });      

    Map<K, V> sortedMap = new LinkedHashMap<K, V>();      

    for (Map.Entry<K, V> entry : entries)
    {
        sortedMap.put(entry.getKey(), entry.getValue());

    }      

    return sortedMap;

}

http://javawithswaranga.blogspot.com/2011/06/generic-method-to-sort-hashmap.html

答案 34 :(得分:2)

当我面对这个时,我只是在旁边创建一个列表。如果你把它们放在一个自定义的Map实现中,它会有一个很好的感觉...你可以使用类似下面的东西,只在需要时执行排序。 (注意:我还没有真正测试过这个,但它编译......在某处可能是一个愚蠢的小虫子)

(如果你希望它按键和值排序,让类扩展TreeMap,不要定义访问器方法,并让mutators调用super.xxxxx而不是map_.xxxx)

package com.javadude.sample;

import java.util.ArrayList;
import java.util.Collection;
import java.util.Collections;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Set;

public class SortedValueHashMap<K, V> implements Map<K, V> {
    private Map<K, V> map_ = new HashMap<K, V>();
    private List<V> valueList_ = new ArrayList<V>();
    private boolean needsSort_ = false;
    private Comparator<V> comparator_;

    public SortedValueHashMap() {
    }
    public SortedValueHashMap(List<V> valueList) {
        valueList_ = valueList;
    }

    public List<V> sortedValues() {
        if (needsSort_) {
            needsSort_ = false;
            Collections.sort(valueList_, comparator_);
        }
        return valueList_;
    }

    // mutators
    public void clear() {
        map_.clear();
        valueList_.clear();
        needsSort_ = false;
    }

    public V put(K key, V value) {
        valueList_.add(value);
        needsSort_ = true;
        return map_.put(key, value);
    }

    public void putAll(Map<? extends K, ? extends V> m) {
        map_.putAll(m);
        valueList_.addAll(m.values());
        needsSort_ = true;
    }

    public V remove(Object key) {
        V value = map_.remove(key);
        valueList_.remove(value);
        return value;
    }

    // accessors
    public boolean containsKey(Object key)           { return map_.containsKey(key); }
    public boolean containsValue(Object value)       { return map_.containsValue(value); }
    public Set<java.util.Map.Entry<K, V>> entrySet() { return map_.entrySet(); }
    public boolean equals(Object o)                  { return map_.equals(o); }
    public V get(Object key)                         { return map_.get(key); }
    public int hashCode()                            { return map_.hashCode(); }
    public boolean isEmpty()                         { return map_.isEmpty(); }
    public Set<K> keySet()                           { return map_.keySet(); }
    public int size()                                { return map_.size(); }
    public Collection<V> values()                    { return map_.values(); }
}

答案 35 :(得分:1)

用于sortHashMap的最简单的强行HashMap<String, Long>方法:您可以复制并使用,如下所示:

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;
    }
}

答案 36 :(得分:1)

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

创建新课程

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());
    }
}

答案 37 :(得分:1)

提供地图

   Map<String, Integer> wordCounts = new HashMap<>();
    wordCounts.put("USA", 100);
    wordCounts.put("jobs", 200);
    wordCounts.put("software", 50);
    wordCounts.put("technology", 70);
    wordCounts.put("opportunity", 200);

根据值升序对地图进行排序

Map<String,Integer>  sortedMap =  wordCounts.entrySet().
                                                stream().
                                                sorted(Map.Entry.comparingByValue()).
        collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));
    System.out.println(sortedMap);

根据值以降序对地图进行排序

Map<String,Integer>  sortedMapReverseOrder =  wordCounts.entrySet().
            stream().
            sorted(Map.Entry.comparingByValue(Comparator.reverseOrder())).
            collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue, (e1, e2) -> e1, LinkedHashMap::new));
    System.out.println(sortedMapReverseOrder);

输出:

{软件= 50,技术= 70,美国= 100,工作= 200,机会= 200}

{职位= 200,机会= 200,美国= 100,技术= 70,软件= 50}

答案 38 :(得分:1)

这样做还有一个好处,就是可以使用Java 8进行升序或降序排序

Uncaught Error: Display(...): Nothing was returned from render. This usually means a return statement is missing. Or, to render nothing, return null.

答案 39 :(得分:1)

在Java 8及更高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及更低版本

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());
}

答案 40 :(得分:0)

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);

答案 41 :(得分:0)

为每个值创建一个条目列表,对值进行排序
需要Java 8或更高版本

Map<Double,List<Entry<String,Double>>> sorted =
map.entrySet().stream().collect( Collectors.groupingBy( Entry::getValue, TreeMap::new,
    Collectors.mapping( Function.identity(), Collectors.toList() ) ) );

使用地图{[A = 99.5],[B = 67.4],[C = 67.4],[D = 67.3]}
得到{67.3=[D=67.3], 67.4=[B=67.4, C=67.4], 99.5=[A=99.5]}


…以及如何依次访问每个条目:

sorted.entrySet().forEach( e -> e.getValue().forEach( l -> System.out.println( l ) ) );

D=67.3 B=67.4 C=67.4 A=99.5

答案 42 :(得分:0)

使用Guava库:

{
  :id=>"AQMkADAwATMwMAItMjE5ZC01YjYyLTAwAi0wMAoARgAAA1D6u4KwtRdMm1rAZoOEKvAHAMf0BP7S_89KmnrxCgze9tcAAAIBBgAAAMf0BP7S_89KmnrxCgze9tcAAAIjxAAAAA==",
  :name=>"Calendar",
  :color=>"auto", 
  :changeKey=>"I8E7t/aNjEqDHtJJL1UC6AAADFI=", 
  :canShare=>true, 
  :canViewPrivateItems=>true, 
  :canEdit=>true, 
  :owner=>{:name=>"User", :address=>"user@outlook.com"}}]
}

答案 43 :(得分:0)

发布我的答案版本

List<Map.Entry<String, Integer>> list = new ArrayList<>(map.entrySet());
    Collections.sort(list, (obj1, obj2) -> obj2.getValue().compareTo(obj1.getValue()));
    Map<String, Integer> resultMap = new LinkedHashMap<>();
    list.forEach(arg0 -> {
        resultMap.put(arg0.getKey(), arg0.getValue());
    });
    System.out.println(resultMap);

答案 44 :(得分:0)

在 Java 中以最简单的方式对任何 Hashmap 进行排序。 我们不需要将它存储在树状图、列表等中。

在这里,我将使用 Java Streams:

让我们按地图的值(升序)对地图进行排序

Map<String, Integer> mp= new HashMap<>();
mp.put("zebra", 1);
mp.put("blossom", 2);
mp.put("gemini", 3);
mp.put("opera", 7);
mp.put("adelaide", 10);

Map<String, Integer> resultMap= mp.entrySet().stream().sorted(Map.Entry.<String, Integer>comparingByValue()).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,(e1, e2) -> e1, LinkedHashMap::new));

您现在可以通过多种方式打印已排序的 resultMap,例如使用高级 for 循环或迭代器。

上图也可以按值降序排列

 Map<String, Integer> resultMap= mp.entrySet().stream().sorted(Map.Entry.<String, Integer>comparingByValue().reversed()).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,(e1, e2) -> e1, LinkedHashMap::new));

现在让我们采取另一个场景,我们将“用户”存储在地图中,并根据“用户”的“名称”按升序(按字典顺序)对其进行排序:

User u1= new User("hi", 135);
User u2= new User("bismuth", 900);
User u3= new User("alloy", 675);
User u4= new User("jupiter", 342);
User u5= new User("lily", 941);

Map<String, User> map2= new HashMap<>();
map2.put("zebra", u3);
map2.put("blossom", u5);
map2.put("gemini", u1);
map2.put("opera", u2);
map2.put("adelaide", u4);


Map<String, User>  resultMap= 
          map2.entrySet().stream().sorted(Map.Entry.<String, User>comparingByValue( (User o1, User o2)-> o1.getName().compareTo(o2.getName()))).collect(Collectors.toMap(Map.Entry::getKey, Map.Entry::getValue,(e1, e2) -> e2, LinkedHashMap::new));



class User
 {
    String name;
    int id;
        

public User(String name, int id) {
    super();
    this.name = name;
    this.id = id;
}
public String getName() {
    return name;
}
public void setName(String name) {
    this.name = name;
}
public int getId() {
    return id;
}
public void setId(int id) {
    this.id = id;
}
@Override
public String toString() {
    return "User [name=" + name + ", id=" + id + "]";
}
@Override
public int hashCode() {
    final int prime = 31;
    int result = 1;
    result = prime * result + id;
    result = prime * result + ((name == null) ? 0 : name.hashCode());
    return result;
}
@Override
public boolean equals(Object obj) {
    if (this == obj)
        return true;
    if (obj == null)
        return false;
    if (getClass() != obj.getClass())
        return false;
    User other = (User) obj;
    if (id != other.id)
        return false;
    if (name == null) {
        if (other.name != null)
            return false;
    } else if (!name.equals(other.name))
        return false;
    return true;


    }
 }

答案 45 :(得分:0)

我可以举一个例子,但是请确保这是您所需要的。

map = {10 = 3, 11 = 1,12 = 2} 

比方说,您想要最频繁使用的前2个键是(10,12) 因此,最简单的方法是使用PriorityQueue根据地图的值进行排序。

PriorityQueue<Integer> pq = new PriorityQueue<>((a, b) -> (map.get(a) - map.get(b));
for(int key: map.keySets()) {
   pq.add(key);
   if(pq.size() > 2) {
      pq.poll();
   }
}
// Now pq has the top 2 most frequent key based on value. It sorts the value. 

答案 46 :(得分:0)

我重写了devinmoore的方法,该方法无需使用Iterator就可以按其值对地图进行排序:

public static Map<K, V> sortMapByValue(Map<K, V> inputMap) {

    Set<Entry<K, V>> set = inputMap.entrySet();
    List<Entry<K, V>> list = new ArrayList<Entry<K, V>>(set);

    Collections.sort(list, new Comparator<Map.Entry<K, V>>()
    {
        @Override
        public int compare(Entry<K, V> o1, Entry<K, V> o2) {
            return (o1.getValue()).compareTo( o2.getValue() );  //Ascending order
        }
    } );

    Map<K, V> sortedMap = new LinkedHashMap<>();

    for(Map.Entry<K, V> entry : list){
        sortedMap.put(entry.getKey(), entry.getValue());
    }

    return sortedMap;
}

Note:,我们将LinkedHashMap用作输出映射,因为我们的列表已按值排序,现在我们应该将列表按插入键,值的顺序存储到输出映射中。因此,如果您使用TreeMap作为输出地图,则地图将再次按地图键排序!

这是主要方法:

public static void main(String[] args) {
    Map<String, String> map = new HashMap<>();
    map.put("3", "three");
    map.put("1", "one");
    map.put("5", "five");
    System.out.println("Input Map:" + map);
    System.out.println("Sorted Map:" + sortMapByValue(map));
}

最后,这是输出:

Input Map:{1=one, 3=three, 5=five}
Sorted Map:{5=five, 1=one, 3=three}

答案 47 :(得分:0)

如果偏爱使用Map数据结构,该数据结构固有地按值排序而不需要触发任何排序方法或显式传递给实用程序,则可以采用以下解决方案:

(1)org.drools.chance.core.util.ValueSortedMap(JBoss项目)在内部维护两个映射,一个用于查找,另一个用于维护排序的值。与以前添加的答案非常相似,但可能是抽象和封装部分(包括复制机制)使从外部使用更安全。

(2)http://techblog.molindo.at/2008/11/java-map-sorted-by-value.html避免维护两个映射,而是依靠Apache Common的LinkedMap进行扩展。 (博客作者注:as all the code here is in the public domain):

// required to access LinkEntry.before and LinkEntry.after
package org.apache.commons.collections.map;

// SNIP: imports

/**
* map implementation based on LinkedMap that maintains a sorted list of
* values for iteration
*/
public class ValueSortedHashMap extends LinkedMap {
    private final boolean _asc;

    // don't use super()!
    public ValueSortedHashMap(final boolean asc) {
        super(DEFAULT_CAPACITY);
        _asc = asc;
    }

    // SNIP: some more constructors with initial capacity and the like

    protected void addEntry(final HashEntry entry, final int hashIndex) {
        final LinkEntry link = (LinkEntry) entry;
        insertSorted(link);
        data[hashIndex] = entry;
    }

    protected void updateEntry(final HashEntry entry, final Object newValue) {
        entry.setValue(newValue);
        final LinkEntry link = (LinkEntry) entry;
        link.before.after = link.after;
        link.after.before = link.before;
        link.after = link.before = null;
        insertSorted(link);
    }

    private void insertSorted(final LinkEntry link) {
        LinkEntry cur = header;
        // iterate whole list, could (should?) be replaced with quicksearch
        // start at end to optimize speed for in-order insertions
        while ((cur = cur.before) != header & amp; & amp; !insertAfter(cur, link)) {}
        link.after = cur.after;
        link.before = cur;
        cur.after.before = link;
        cur.after = link;
    }

    protected boolean insertAfter(final LinkEntry cur, final LinkEntry link) {
        if (_asc) {
            return ((Comparable) cur.getValue())
            .compareTo((V) link.getValue()) & lt; = 0;
        } else {
            return ((Comparable) cur.getValue())
            .compareTo((V) link.getValue()) & gt; = 0;
        }
    }

    public boolean isAscending() {
        return _asc;
    }
}

(3)编写自定义Map或从LinkedHashMap扩展而来,仅在枚举期间进行排序(例如values()keyset()entryset()),如下所示:需要。内部实现/行为是使用此类从一个实现中抽象出来的,但是对于此类的客户端而言,似乎要求枚举时总是对值进行排序。此类希望在枚举之前完成所有put操作的情况下,排序几乎只会发生一次。排序方法采用了该问题的先前答案。

public class SortByValueMap<K, V> implements Map<K, V> {

    private boolean isSortingNeeded = false;

    private final Map<K, V> map = new LinkedHashMap<>();

    @Override
    public V put(K key, V value) {
        isSortingNeeded = true;
        return map.put(key, value);
    }

    @Override
    public void putAll(Map<? extends K, ? extends V> map) {
        isSortingNeeded = true;
        map.putAll(map);
    }

    @Override
    public Set<K> keySet() {
        sort();
        return map.keySet();
    }

    @Override
    public Set<Entry<K, V>> entrySet() {
        sort();
        return map.entrySet();
    }

    @Override
    public Collection<V> values() {
        sort();
        return map.values();
    }

    private void sort() {
        if (!isSortingNeeded) {
            return;
        }

        List<Entry<K, V>> list = new ArrayList<>(size());

        for (Iterator<Map.Entry<K, V>> it = map.entrySet().iterator(); it.hasNext();) {
            Map.Entry<K, V> entry = it.next();
            list.add(entry);
            it.remove();
        }

        Collections.sort(list);

        for (Entry<K, V> entry : list) {
            map.put(entry.getKey(), entry.getValue());
        }

        isSortingNeeded = false;
    }

    @Override
    public String toString() {
        sort();
        return map.toString();
    }
}

(4)Guava提供了ImmutableMap.Builder.orderEntriesByValue(Comparator valueComparator),尽管生成的地图将是不可变的:

  

配置此构建器以根据值按值对条目进行排序   指定的比较器。

     

排序顺序是稳定的,也就是说,如果两个条目的值分别为   比较等效,首先插入的条目将是第一个   按构建地图的迭代顺序。

答案 48 :(得分:0)

我的解决方案是一种使用大多数给定API的方法。 我们使用地图的功能,通过 entrySet()方法将其内容导出为设置。我们现在有一个包含 Map.Entry 对象的 Set

好的,Set没有订单,但我们可以将内容放入 ArrayList 。它现在有随机顺序,但无论如何我们都会对它进行排序。

由于 ArrayList 集合,我们现在使用 Collections.sort()方法将秩序置于混乱状态。由于我们的 Map.Entry 对象没有实现我们需要的那种比较,因此我们提供了自定义比较器

public static void main(String[] args) {
    HashMap<String, String> map = new HashMap<>();
    map.put("Z", "E");
    map.put("G", "A");
    map.put("D", "C");
    map.put("E", null);
    map.put("O", "C");
    map.put("L", "D");
    map.put("Q", "B");
    map.put("A", "F");
    map.put(null, "X");
    MapEntryComparator mapEntryComparator = new MapEntryComparator();

    List<Entry<String,String>> entryList = new ArrayList<>(map.entrySet());
    Collections.sort(entryList, mapEntryComparator);

    for (Entry<String, String> entry : entryList) {
        System.out.println(entry.getKey() + " : " + entry.getValue());
    }

}

答案 49 :(得分:-1)

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);
    }   
}

OUTPUT =

//sort by values 

3 arun singh
1 bandagi singh
28 jakarta singh
2 panipat singh
6 vikram singh
5 vinay singh

答案 50 :(得分:-1)

为了对键进行排序,我找到了一个更好的TreeMap解决方案(我将尝试为基于值的排序提供解决方案):

public static void main(String[] args) {
    Map<String, String> unsorted = new HashMap<String, String>();
    unsorted.put("Cde", "Cde_Value");
    unsorted.put("Abc", "Abc_Value");
    unsorted.put("Bcd", "Bcd_Value");

    Comparator<String> comparer = new Comparator<String>() {
        @Override
        public int compare(String o1, String o2) {
            return o1.compareTo(o2);
        }};

    Map<String, String> sorted = new TreeMap<String, String>(comparer);
    sorted.putAll(unsorted);
    System.out.println(sorted);
}

输出将是:

{Abc = Abc_Value,Bcd = Bcd_Value,Cde = Cde_Value}

答案 51 :(得分:-1)

public class SortedMapExample {

    public static void main(String[] args) {
        Map<String, String> map = new HashMap<String, String>();

        map.put("Cde", "C");
        map.put("Abc", "A");
        map.put("Cbc", "Z");
        map.put("Dbc", "D");
        map.put("Bcd", "B");
        map.put("sfd", "Bqw");
        map.put("DDD", "Bas");
        map.put("BGG", "Basd");

        System.out.println(sort(map, new Comparator<String>() {
            @Override
            public int compare(String o1, String o2) {
                    return o1.compareTo(o2);
            }}));
    }

    @SuppressWarnings("unchecked")
    public static <K, V> Map<K,V> sort(Map<K, V> in, Comparator<? super V> compare) {
        Map<K, V> result = new LinkedHashMap<K, V>();
        V[] array = (V[])in.values().toArray();
        for(int i=0;i<array.length;i++)
        {

        }
        Arrays.sort(array, compare);
        for (V item : array) {
            K key= (K) getKey(in, item);
            result.put(key, item);
        }
        return result;
    }

    public static <K, V>  Object getKey(Map<K, V> in,V value)
    {
       Set<K> key= in.keySet();
       Iterator<K> keyIterator=key.iterator();
       while (keyIterator.hasNext()) {
           K valueObject = (K) keyIterator.next();
           if(in.get(valueObject).equals(value))
           {
                   return valueObject;
           }
       }
       return null;
   }

}

//请在这里试试我正在修改值排序的代码。

答案 52 :(得分:-2)

    static <K extends Comparable<? super K>, V extends Comparable<? super V>>
    Map sortByValueInDescendingOrder(final Map<K, V> map) {
        Map re = new TreeMap(new Comparator<K>() {
            @Override
            public int compare(K o1, K o2) {
                if (map.get(o1) == null || map.get(o2) == null) {
                    return -o1.compareTo(o2);
                }
                int result = -map.get(o1).compareTo(map.get(o2));
                if (result != 0) {
                    return result;
                }
                return -o1.compareTo(o2);
            }
        });
        re.putAll(map);
        return re;
    }
    @Test(timeout = 3000l, expected = Test.None.class)
    public void testSortByValueInDescendingOrder() {
        char[] arr = "googler".toCharArray();
        Map<Character, Integer> charToTimes = new HashMap();
        for (int i = 0; i < arr.length; i++) {
            Integer times = charToTimes.get(arr[i]);
            charToTimes.put(arr[i], times == null ? 1 : times + 1);
        }
        Map sortedByTimes = sortByValueInDescendingOrder(charToTimes);
        Assert.assertEquals(charToTimes.toString(), "{g=2, e=1, r=1, o=2, l=1}");
        Assert.assertEquals(sortedByTimes.toString(), "{o=2, g=2, r=1, l=1, e=1}");
        Assert.assertEquals(sortedByTimes.containsKey('a'), false);
        Assert.assertEquals(sortedByTimes.get('a'), null);
        Assert.assertEquals(sortedByTimes.get('g'), 2);
        Assert.assertEquals(sortedByTimes.equals(charToTimes), true);
    }

答案 53 :(得分:-2)

如果没有任何大于地图大小的值,你可以使用数组,这应该是最快的方法:

public List<String> getList(Map<String, Integer> myMap) {
    String[] copyArray = new String[myMap.size()];
    for (Entry<String, Integer> entry : myMap.entrySet()) {
        copyArray[entry.getValue()] = entry.getKey();
    }
    return Arrays.asList(copyArray);
}

答案 54 :(得分:-2)

好的,这个版本适用于两个新的Map对象和两个迭代并对值进行排序。希望,尽管地图条目必须循环两次,但表现还是不错的:

public static void main(String[] args) {
    Map<String, String> unsorted = new HashMap<String, String>();
    unsorted.put("Cde", "Cde_Value");
    unsorted.put("Abc", "Abc_Value");
    unsorted.put("Bcd", "Bcd_Value");

    Comparator<String> comparer = new Comparator<String>() {
        @Override
        public int compare(String o1, String o2) {
            return o1.compareTo(o2);
        }};

    System.out.println(sortByValue(unsorted, comparer));

}

public static <K, V> Map<K,V> sortByValue(Map<K, V> in, Comparator<? super V> compare) {
    Map<V, K> swapped = new TreeMap<V, K>(compare);
    for(Entry<K,V> entry: in.entrySet()) {
        if (entry.getValue() != null) {
            swapped.put(entry.getValue(), entry.getKey());
        }
    }
    LinkedHashMap<K, V> result = new LinkedHashMap<K, V>();
    for(Entry<V,K> entry: swapped.entrySet()) {
        if (entry.getValue() != null) {
            result.put(entry.getValue(), entry.getKey());
        }
    }
    return result;
}

该解决方案使用带有比较器的TreeMap并对所有空键和值进行排序。首先,TreeMap的排序功能用于对值进行排序,接下来,排序的Map用于创建结果,因为LinkedHashMap保留的值具有相同的值。

Greetz,GHad

答案 55 :(得分:-2)

我们只是按照这样排序地图

            Map<String, String> unsortedMap = new HashMap<String, String>();

    unsortedMap.put("E", "E Val");
    unsortedMap.put("F", "F Val");
    unsortedMap.put("H", "H Val");
    unsortedMap.put("B", "B Val");
    unsortedMap.put("C", "C Val");
    unsortedMap.put("A", "A Val");
    unsortedMap.put("G", "G Val");
    unsortedMap.put("D", "D Val");

    Map<String, String> sortedMap = new TreeMap<String, String>(unsortedMap);

    System.out.println("\nAfter sorting..");
    for (Map.Entry <String, String> mapEntry : sortedMap.entrySet()) {
        System.out.println(mapEntry.getKey() + " \t" + mapEntry.getValue());

答案 56 :(得分:-4)

最好的办法是将HashMap转换为TreeMap。 TreeMap自己对键进行排序。 如果您希望对值进行排序而不是快速修复,则可以使用键切换值,如果您的值不重复。

答案 57 :(得分:-8)

因为地图是无序的 为了排序,我们可以做以下

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。

答案 58 :(得分:-9)

如果您的Map值实现了Comparable(例如String),那么这应该可以工作

Map<Object, String> map = new HashMap<Object, String>();
// Populate the Map
List<String> mapValues = new ArrayList<String>(map.values());
Collections.sort(mapValues);

如果地图值本身没有实现Comparable,但是你有一个可以对它们进行排序的Comparable实例,则将最后一行替换为:

Collections.sort(mapValues, comparable);

答案 59 :(得分:-11)

使用 java.util.TreeMap

“地图是根据其键的自然顺序排序的,或者是在地图创建时提供的比较器,具体取决于使用的构造函数。”