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K Nearest Neighbor问题的解决——KD-TREE Implementation

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命题一:
已知的1000个整数的数组,给定一个整数,要求查证是否在数组中出现?

命题二:
已知1000个整数的数组,给定一个整数,要求查找数组中与之最接近的数字?

命题三:
已知1000个Point(包含X与Y坐标)结构的数组,给定一个Point,要求查找数组中与之最接近(比如:欧氏距离最短)的点。

命题四:
已知1,000,000个向量,每个向量为128维;给定一个向量,要求查找数组中与之最接近的K个向量

  • 对于命题一,如果不考虑桶式、哈希等方式,常用的方法应该是排序后,使用折半查找。
  • 对于命题二,与命题一类似,比较折半查找得出的结果,以及附近的各一个元素,即可。整个过程相当于是把这个包含1000个数组的数据结构做成一颗二叉树,最后只需比较叶子节点与其父节点即可。
  • 对于命题三、四其中命题三和四就是所谓的Nearest Neighbor问题。一种近似解决的方法就是KD-TREE


高维向量的KNN检索问题,在图像等多媒体内容搜索中是相当关键的。关于高维向量的讨论,网上资料比较少;在此,我将一些心得分享给大家。
与二叉树相比,KD-TREE也采用类似的划分方式,只不过树中的各节点均是高维向量,因此划分的方式,采用随机或指定的方式选取一个维度,在该指定维度上进行划分;整体的思想就是采用多个超平面对数据集空间进行两两切分,这一点,有点类似于数据挖掘中的决策树。

一个运用KD-TREE分割二维平面的DEMO如下:



KD-Tree build的代码如下:
private ClusterKDTree(Clusterable[] points, int height, boolean randomSplit){
    if ( points.length == 1 ){
        cluster = points[0];
    }
    else {
        splitIndex = chooseSplitDimension//选取切分维度
            (points[0].getLocation().length,height,randomSplit);
        splitValue = chooseSplit(points,splitIndex);//选取切分值
			
        Vector<Clusterable> left = new Vector<Clusterable>();
        Vector<Clusterable> right = new Vector<Clusterable>();
        for ( int i = 0; i < points.length; i++ ){
            double val = points[i].getLocation()[splitIndex];
            if ( val == splitValue && cluster == null ){
                cluster = points[i];
            }
            else if ( val >= splitValue ){
                right.add(points[i]);
            } else {
                left.add(points[i]);
            }
        }
			
        if ( right.size() > 0 ){
            this.right = new ClusterKDTree(right.toArray(new
            Clusterable[right.size()]),
            randomSplit ? splitIndex : height+1, randomSplit);
        }
        if ( left.size() > 0 ){
            this.left = new ClusterKDTree(left.toArray(new
            Clusterable[left.size()]),randomSplit ? splitIndex : height+1,
            randomSplit);
        }
    }
}

private int chooseSplitDimension(int dimensionality,int height,boolean random){
    if ( !random ) return height % dimensionality;
    int rand = r.nextInt(dimensionality);
    while ( rand == height ){
        rand = r.nextInt(dimensionality);
    }
    return rand;
}
	
private double chooseSplit(Clusterable points[],int splitIdx){
    double[] values = new double[points.length];
    for ( int i = 0; i < points.length; i++ ){
	values[i] = points[i].getLocation()[splitIdx];
    }
    Arrays.sort(values);
    return values[values.length/2];//选取中间值以保持树的平衡
}


构建完一颗KD-TREE之后,如何使用它来做KNN检索呢?我用下面的图来表示(20s的GIF动画):



使用KD-TREE,经过一次二分查找可以获得Query的KNN(最近邻)贪心解,代码如下:
private Clusterable restrictedNearestNeighbor(Clusterable point, SizedPriorityQueue<ClusterKDTree> values){
    if ( splitIndex == -1 ) {
        return cluster; //已近到叶子节点
    }
		
    double val = point.getLocation()[splitIndex];
    Clusterable closest = null;
    if ( val >= splitValue && right != null || left == null ){
        //沿右边路径遍历,并将左边子树放进队列
        if ( left != null ){
            double dist = val - splitValue;
            values.add(left,dist);
        }
        closest = right.restrictedNearestNeighbor(point,values);
    }
    else if ( val < splitValue && left != null || right == null ) {
        //沿左边路径遍历,并将右边子树放进队列
        if ( right != null ){
            double dist = splitValue - val;
            values.add(right,dist);
        }
        closest = left.restrictedNearestNeighbor(point,values);
    }
    //current distance of the 'ideal' node
    double currMinDistance = ClusterUtils.getEuclideanDistance(closest,point);
    //check to see if the current node we've backtracked to is closer
    double currClusterDistance = ClusterUtils.getEuclideanDistance(cluster,point);
    if ( closest == null || currMinDistance > currClusterDistance ){
        closest = cluster;
        currMinDistance = currClusterDistance;
    }
    return closest;
}


事实上,仅仅一次的遍历会有不小的误差,因此采用了一个优先级队列来存放每次决定遍历走向时,另一方向的节点。SizedPriorityQueue代码的实现,可参考我的另一篇文章:
http://grunt1223.iteye.com/blog/909739

一种减少误差的方法(BBF:Best Bin First)是回溯一定数量的节点:
public Clusterable restrictedNearestNeighbor(Clusterable point, int numMaxBinsChecked){
    SizedPriorityQueue<ClusterKDTree> bins = new SizedPriorityQueue<ClusterKDTree>(50,true);
    Clusterable closest = restrictedNearestNeighbor(point,bins);
    double closestDist = ClusterUtils.getEuclideanDistance(point,closest);
    //System.out.println("retrieved point: " + closest + ", dist: " + closestDist);
    int count = 0;
    while ( count < numMaxBinsChecked && bins.size() > 0 ){
        ClusterKDTree nextBin = bins.pop();
	//System.out.println("Popping of next bin: " + nextBin);
	Clusterable possibleClosest = nextBin.restrictedNearestNeighbor(point,bins);
        double dist = ClusterUtils.getEuclideanDistance(point,possibleClosest);
        if ( dist < closestDist ){
	    closest = possibleClosest;
	    closestDist = dist;
	}
	count++;
    }
    return closest;
}


可以用如下代码进行测试:
public static void main(String args[]){
    Clusterable clusters[] = new Clusterable[10];
    clusters[0] = new Point(0,0);
    clusters[1] = new Point(1,2);
    clusters[2] = new Point(2,3);
    clusters[3] = new Point(1,5);
    clusters[4] = new Point(2,5);
    clusters[5] = new Point(1,1);
    clusters[6] = new Point(3,3);
    clusters[7] = new Point(0,2);
    clusters[8] = new Point(4,4);
    clusters[9] = new Point(5,5);
    ClusterKDTree tree = new ClusterKDTree(clusters,true);
    //tree.print();
    Clusterable c = tree.restrictedNearestNeighbor(new Point(4,4),1000);
    System.out.println(c);
}
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评论
1 楼 timedcy 2013-10-13  
Hi, 你这个20s的gif图是自己画的吗?

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