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java - Map基于Value值排序

转载 作者:行者123 更新时间:2023-11-30 07:08:26 24 4
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方法1:

使用TreeMap,可以参考下面的代码

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
}
}

译注:如果不自己写Comparator,treemap默认是用key来排序

方法2:

先通过linkedlist排好序,再放到LinkedHashMap中

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;
}
}

译注:这两种方法,我简单测试了下,如果map的size在十万级别以上,两者的耗时都是几百毫秒,第二个方法会快一些。否则,第一个方法快一些。因此,如果你处理的map,都是几十万级别以下的大小,两种方式随意使用,看个人喜欢了。

stackoverflow链接:http://stackoverflow.com/questions/109383/how-to-sort-a-mapkey-value-on-the-values-in-java

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