我的Hash地图中有两个数组,我想根据averageValueArray
中的时间对存储在timeStampArray
中的值进行排序。尝试使用TreeMap,但弄得一团糟。
任何帮助都会非常明显。
Map<List<Date>,List<Double>> unsortedMap = new HashMap<List<Date>,List<Double>>();
unsortedMap.put(timeStampArray, averageValueArray);
这就是我正在尝试的
Map<List<Date>,List<Double>> sortMap = new HashMap<List<Date>,List<Double>>();
sortMap.put(timeStampArray, averageValueArray);
for (Map.Entry entry : sortMap.entrySet()) {
System.out.println("Key = " + entry.getKey());
System.out.println(" Value = " +entry.getValue());
}
System.out.println("Unsort Map......");
printMap(sortMap);
System.out.println("Sorted Map......");
Map<List<Date>,List<Double>> treeMap = new TreeMap<List<Date>,List<Double>>(sortMap);
printMap(treeMap);
printMap as:
public static void printMap(Map<List<Date>,List<Double>> map) {
for (Map.Entry entry : map.entrySet()) {
System.out.println("Key : " + entry.getKey() + " Value : "
+ entry.getValue());}}
答案 0 :(得分:4)
如果我理解正确,你有两个并行列表,一个包含时间,另一个包含平均值。你希望这两个列表“并行”排序。
你最好有一个对象列表,每个对象包含一个日期和一个平均值,并根据需要对该列表进行排序:
public final class DatedAverage {
private final Date date;
private final double average;
// constructor, getters omitted
}
...
List<DatedAverage> datedAverages = ...;
Collections.sort(datedAverages, new Comparator<DatedAverage>() {
@Override
public int compare(DatedAverage d1, DatedAverage d2) {
return d1.getDate().compareTo(d2.getDate());
}
});
Java是一种OO语言。使用对象,并在这些对象中封装行为。
答案 1 :(得分:0)
我建议您将两个列表合并为一个对象。您可以按照原样保留地图,然后使用此组合列表进行排序,使用下面的静态方法。
public class AverageValueTimeStamp implements Comparable<AverageValueTimeStamp>{
Date timeStamp;
double averageValue;
public AverageValueTimeStamp(Date when, double avg) {
timeStamp = when;
averageValue = avg;
}
public int compareTo(AverageValueTimeStamp other) {
if(timeStamp.equals(other.timeStamp)
retrn averageValue - other.AverageValue;
else
return timeStamp.compareTo(other.timeStamp);
}
/**
* @return a list of AverageValueTimeStamp, sorted by Date (timestamp).
*/
public static ArrayList<AverageValueTimeStamp> toList(List<Date> timeStamps, List<Double> averages) {
//change to iterators if these might be LinkedLists
ArrayList<AverageValueTimeStamp> avtsList = new ArrayList<>( timeStamps.size() );
for(int i=0; i<timeStamps.size(); i++) {
AverageValueTimeStamp avts = new AverageValueTimeStamp(timeStamps.get(i), averages.get(i));
avtsList.add(avts);
}
Collections.sort(avtsList);
return avtsList;
}
}
答案 2 :(得分:0)
我建议你单独处理2个列表,它看起来像这样
public HashMap sortLists(List list1, List list2){ HashMap,List> map = new HashMap,List>(); Collections.sort(list1); //sort the date list ArrayList sorted = new ArrayList(); for(Date date : list1){ //your logic goes here, add objects to sorted //use this method when iterating your hasmap for each key value //if you want return the sorted list instead of hashmap } map.put(list1, sorted); return map; }