我一直在想它，但已经用完了。我有 10 个数组，每个数组的长度为 18，其中有 18 个 double 值。这 18 个值是图像的特征。现在我必须对它们应用 k-means 聚类。

为了实现 k 均值聚类，我需要为每个数组提供唯一的计算值。是否有任何数学或统计或任何逻辑可以帮助我为每个数组创建一个计算值，它基于其中的值是唯一的。提前致谢。

这是我的数组示例。还有10个

[0.07518284315321135    
0.002987851573676068    
0.002963866526639678    
0.002526139418225552    
0.07444872939213325 
0.0037219653347541617   
0.0036979802877177715   
0.0017920256571474585   
0.07499695903867931 
0.003477831820276616    
0.003477831820276616    
0.002036159171625004    
0.07383539747505984 
0.004311312204791184    
0.0043352972518275745   
0.0011786937400740452   
0.07353130134299131 
0.004339580295941216]

最佳答案

您检查过 Java 7 中的 Arrays.hashcode 了吗？

 /**
 * Returns a hash code based on the contents of the specified array.
 * For any two <tt>double</tt> arrays <tt>a</tt> and <tt>b</tt>
 * such that <tt>Arrays.equals(a, b)</tt>, it is also the case that
 * <tt>Arrays.hashCode(a) == Arrays.hashCode(b)</tt>.
 *
 * <p>The value returned by this method is the same value that would be
 * obtained by invoking the {@link List#hashCode() <tt>hashCode</tt>}
 * method on a {@link List} containing a sequence of {@link Double}
 * instances representing the elements of <tt>a</tt> in the same order.
 * If <tt>a</tt> is <tt>null</tt>, this method returns 0.
 *
 * @param a the array whose hash value to compute
 * @return a content-based hash code for <tt>a</tt>
 * @since 1.5
 */
public static int hashCode(double a[]) {
    if (a == null)
        return 0;

    int result = 1;
    for (double element : a) {
        long bits = Double.doubleToLongBits(element);
        result = 31 * result + (int)(bits ^ (bits >>> 32));
    }
    return result;
}

我不明白为什么@Marco13 提到“这不会为数组返回异常”。

更新

请参阅@Macro13 对 the reason 的评论为什么它不能不安静..

更新

如果我们使用您的输入点绘制图形，(18 个元素)有一个尖峰值和 3 个低值，并且该模式为…… 如果那是真的..你可以找到你的峰值(1、4、8、12、16)的平均值，并从剩余的值中找到低平均值。

因此您将拥有 Peak mean 和 Low mean 。并且您会发现表示这两个的异常数也使用 here 中描述的双射算法保留值

此 Alogirthm 还提供了反向公式，即从非静止值中获取峰值和低均值。

寻找唯一对 < x; y >= x + (y + ( (( x +1 ) /2) * (( x +1 ) /2) ) )

另请参阅 pdf 第 2 页中的练习 1 以反转 x 和 y。

用于求均值和配对值。

public static double mean(double[] array){
    double peakMean = 0;
    double lowMean = 0;
    for (int i = 0; i < array.length; i++) {
        if ( (i+1) % 4 == 0 || i == 0){
            peakMean = peakMean + array[i];
        }else{
            lowMean = lowMean + array[i];
        }
    }
    peakMean = peakMean / 5;
    lowMean = lowMean / 13;
    return bijective(lowMean, peakMean);
}



public static double bijective(double x,double y){
    double tmp = ( y +  ((x+1)/2));
    return x +  ( tmp * tmp);
}

测试

public static void main(String[] args) {
    double[] arrays = {0.07518284315321135,0.002963866526639678,0.002526139418225552,0.07444872939213325,0.0037219653347541617,0.0036979802877177715,0.0017920256571474585,0.07499695903867931,0.003477831820276616,0.003477831820276616,0.002036159171625004,0.07383539747505984,0.004311312204791184,0.0043352972518275745,0.0011786937400740452,0.07353130134299131,0.004339580295941216};
    System.out.println(mean(arrays));
}

您可以使用此峰值和低值来查找相似图像。

关于java - 数组的唯一计算值，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/22342898/