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
用Java中最简单的方式对任何Hashmap进行排序。我们不需要将其存储在树图、列表等中。
在这里,我将使用Java Streams:
让我们按其值(升序)对该地图进行排序
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));
现在,您可以通过多种方式打印排序后的resultMap,例如使用高级for循环或迭代器。
上面的映射也可以按值的降序排序
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));
现在让我们来看另一个场景,我们将“User”存储在地图中,并根据“User”的“name”按升序(词典)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;
}
}
其他回答
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
创建新类
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());
}
}
当我面对这个问题时,我只是在旁边创建一个列表。如果您将它们放在一个自定义的Map实现中,它会有一种很好的感觉……您可以使用类似以下的方式,仅在需要时执行排序。(注意:我还没有真正测试过这个,但它可以编译……可能是某个地方的一个愚蠢的小bug)
(如果您希望按键和值对其进行排序,请让类扩展TreeMap,不要定义访问器方法,并让赋值函数调用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(); }
}
最干净的方法是利用集合对值进行排序:
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());
}
}
基于@devinmore代码,一种使用泛型并支持升序和降序排序的map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
}
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()
);
}