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
在TreeMap中,键按自然顺序排序。例如,如果您对数字进行排序,(注意4的排序)
{0=0, 10=10, 20=20, 30=30, 4=4, 50=50, 60=60, 70=70}
要解决这个问题,在Java8中,首先检查字符串长度,然后进行比较。
Map<String, String> sortedMap = new TreeMap<>Comparator.comparingInt(String::length)
.thenComparing(Function.identity()));
{0=0, 4=4, 10=10, 20=20, 30=30, 50=50, 60=60, 70=70}
其他回答
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);
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());
}
}
对键进行排序需要Comparator为每个比较查找每个值。一个更具可扩展性的解决方案将直接使用entrySet,因为这样每次比较都会立即获得该值(尽管我没有用数字来支持)。
这是这样一件事的通用版本:
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());
}
}
有一些方法可以减少上述解决方案的内存旋转。例如,创建的第一个ArrayList可以重新用作返回值;这将需要抑制一些泛型警告,但对于可重用的库代码来说,这可能是值得的。此外,Comparator不必在每次调用时重新分配。
这里有一个更有效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,如果您需要连续访问已排序的信息(而不是偶尔排序一次),可以使用额外的多重映射。如果你需要更多细节,请告诉我。。。
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
}
主要问题。如果您使用第一个答案(Google将您带到这里),请更改比较器以添加等号子句,否则无法按键从sorted_map中获取值:
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
}