Java 如何使用LinkedHashMap获取submap?
目前,我正在使用Java 如何使用LinkedHashMap获取submap?,java,performance,dictionary,iteration,Java,Performance,Dictionary,Iteration,目前,我正在使用TreeMap存储一些x和y坐标,但与ArrayList或HashMap相比,迭代非常缓慢。我之所以使用它,是因为我需要subMap()方法,这样即使精确的X值(键)不存在,我也可以在确定的范围内获得X值 LinkedHashMap的速度几乎与HashMap相同,我可以按插入顺序迭代键(我需要插入顺序或比较器顺序,就像在TreeMap中那样),但我没有submap()方法。在树状图中,我可以快速生成子图 是否有任何数据结构或某种方式可以比TreeMap更快地存储有序值(通过插入顺
TreeMap
存储一些x和y坐标,但与ArrayList
或HashMap
相比,迭代非常缓慢。我之所以使用它,是因为我需要subMap()
方法,这样即使精确的X值(键)不存在,我也可以在确定的范围内获得X值
LinkedHashMap
的速度几乎与HashMap
相同,我可以按插入顺序迭代键(我需要插入顺序或比较器顺序,就像在TreeMap中那样),但我没有submap()
方法。在树状图中,我可以快速生成子图
是否有任何数据结构或某种方式可以比TreeMap更快地存储有序值(通过插入顺序或比较器),即使精确值不在映射中,也可以在一定范围内获取子映射?我的意思是,也许我想要2到25之间的值,但是2不存在,最近的值是3,所以它将返回一个从3到25的子映射。或者以某种方式将此功能添加到
LinkedHashMap
听起来你需要一个树形图,它的迭代速度不比LinkedHashMap慢多少,并且做你真正想要的事情。由于HashMap是无序的,subMap没有任何意义。今天我终于找到了问题的答案。经过几次测试,HashMap
,LinkedHashMap
和TreeMap
比ArrayList
慢得多,我想使用它们只是为了能够创建submap()
。因此,我创建了一个扩展ArrayList
的新类,它给了我一个非常好的性能,在answer的帮助下,我创建了一个快速的方法,通过值而不是索引获取子列表。以下是完整的课程:
/**
* The purpose of this class is to be a faster replacement to a {@link java.util.TreeMap} with
* the ability to get sublist containing a range of x values. ArrayList access time is O(1) while
* {@link java.util.TreeMap} is O(log(n)). When large data is handled the impact on performance is
* noticeable.
*/
public class XYDataset extends ArrayList<PointValue> {
private final float COMPARISON_THRESHOLD = 0.01f;
final Comparator<PointValue> comparator = new Comparator<PointValue>() {
@Override
public int compare(PointValue lhs, PointValue rhs) {
if (Math.abs(lhs.getX() - rhs.getX()) < COMPARISON_THRESHOLD) return 0;
return lhs.getX() < rhs.getX() ? -1 : 1;
}
};
public XYDataset(int capacity) {
super(capacity);
}
public XYDataset() {
}
public XYDataset(Collection<? extends PointValue> collection) {
super(collection);
}
@Override
public List<PointValue> subList(int start, int end) {
return super.subList(start, end);
}
/**
* Generate a sublist containing the range of x values passed
* @param x1 lower x value
* @param x2 upper x value
* @return sublist containing x values from x1 to x2
*/
public List<PointValue> subList(float x1, float x2){
/**
* Collections.binarySearch() returns the index of the search key, if it is contained in the list;
* otherwise it returns (-(insertion point) - 1).
* The insertion point is defined as the point at which the key would be inserted into the list:
* the index of the first element greater than the key, or list.size() if all elements in the list
* are less than the specified key. Note that this guarantees that the return value will be >= 0 if
* and only if the key is found.
*/
int n1 = Collections.binarySearch(this, new PointValue(x1, 0), comparator);
int n2 = Collections.binarySearch(this, new PointValue(x2, 0), comparator);
/**
* Example, we assume the list is sorted. Based on (https://stackoverflow.com/questions/19198586/search-sorted-listlong-for-closest-and-less-than)
*
* long X = 500;
* List<Long> foo = new Arraylist<>();
* foo.add(450L);
* foo.add(451L);
* foo.add(499L);
* foo.add(501L);
* foo.add(550L);
*
* If we search for something that isn't in the list you can work backward from the return value
* to the index you want. If you search for 500 in your example list, the algorithm would return (-3 - 1) = -4.
* Thus, you can add 1 to get back to the insertion point (-3), and then multiply by -1 and subtract 1 to get
* the index BEFORE the first element GREATER than the one you searched for, which will either be an index that
* meets your 2 criteria OR -1 if all elements in the list are greater than the one you searched for.
*/
if(n1 < 0) n1 = -n1-1;
if(n2 < 0) n2 = -n2-1;
return this.subList(n1, n2);
}
}
/**
*这个类的目的是更快地替换{@link java.util.TreeMap},使用
*获取包含x值范围的子列表的能力。ArrayList访问时间为O(1),而
*{@link java.util.TreeMap}是O(log(n))。在处理大数据时,对性能的影响很小
*显而易见。
*/
公共类XYDataset扩展了ArrayList{
私人最终浮动比较_阈值=0.01f;
最终比较器比较器=新比较器(){
@凌驾
公共整数比较(点值lhs、点值rhs){
if(Math.abs(lhs.getX()-rhs.getX()) public XYDataset(CollectionI当前正在使用TreeMap,但与具有O(1)的HashMap或ArrayList相比,它的速度非常慢 access@Andres关注的是访问时间还是迭代时间?是迭代时间,我迭代了500k个条目。但是如果访问时间较低,迭代时间应该较低还是不低?@Andres迭代与随机访问是不同的操作,我不希望它是相同的。