对地图进行排序< Key，Value>按值（Java） [英] Sort a Map<Key, Value> by values (Java)

问题描述

我对Java比较陌生，经常发现我需要对值进行排序 Map< Key，Value> 。因为值不唯一，我发现自己将 keySet 转换为数组，并通过数组排序，并使用自定义比较对与该键相关联的值进行排序。有更简单的方法吗？

I am relatively new to Java, and often find that I need to sort a Map<Key, Value> on the values. Since the values are not unique, I find myself converting the keySet into an array, and sorting that array through array sort with a custom comparator that sorts on the value associated with the key. Is there an easier way?

推荐答案

这是一个友好的版本，你可以免费使用：

Here's a generic-friendly version you're free to use:

import java.util.*;

public class MapUtil
{
    public static <K, V extends Comparable<? super V>> Map<K, V> 
        sortByValue( 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() );
            }
        } );

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

和一个相关的JUnit4测试，以我的话为它：

And an associated JUnit4 test so you don't have to take my word for it:

import java.util.*;
import org.junit.*;

public class MapUtilTest
{
    @Test
    public void testSortByValue()
    {
        Random random = new Random(System.currentTimeMillis());
        Map<String, Integer> testMap = new HashMap<String, Integer>(1000);
        for(int i = 0 ; i < 1000 ; ++i) {
            testMap.put( "SomeString" + random.nextInt(), random.nextInt());
        }

        testMap = MapUtil.sortByValue( testMap );
        Assert.assertEquals( 1000, testMap.size() );

        Integer previous = null;
        for(Map.Entry<String, Integer> entry : testMap.entrySet()) {
            Assert.assertNotNull( entry.getValue() );
            if (previous != null) {
                Assert.assertTrue( entry.getValue() >= previous );
            }
            previous = entry.getValue();
        }
    }

}

版本

Java 7 Version

public static <K, V extends Comparable<? super V>> Map<K, V> 
    sortByValue( Map<K, V> map )
{
    List<Map.Entry<K, V>> list =
        new LinkedList<>( map.entrySet() );
    Collections.sort( list, new Comparator<Map.Entry<K, V>>()
    {
        @Override
        public int compare( Map.Entry<K, V> o1, Map.Entry<K, V> o2 )
        {
            return ( o1.getValue() ).compareTo( o2.getValue() );
        }
    } );

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

Java 8版本。这将根据值按升序排序;对于降序，只需取消对 Collections.reverseOrder（）的调用。

Java 8 Version. This will sort according to the value in ascending order; for descending order, it is just possible to uncomment the call to Collections.reverseOrder().

public static <K, V extends Comparable<? super V>> Map<K, V> sortByValue(Map<K, V> map) {
    return map.entrySet()
              .stream()
              .sorted(Map.Entry.comparingByValue(/*Collections.reverseOrder()*/))
              .collect(Collectors.toMap(
                Map.Entry::getKey, 
                Map.Entry::getValue, 
                (e1, e2) -> e1, 
                LinkedHashMap::new
              ));
}

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