Unique Computational value for an array

Question

I have been thinking of it but have ran out of idea's. I have 10 arrays each of length 18 and having 18 double values in them. These 18 values are features of an image. Now I have to apply k-means clustering on them.

For implementing k-means clustering I need a unique computational value for each array. Are there any mathematical or statistical or any logic that would help me to create a computational value for each array, which is unique to it based upon values inside it. Thanks in advance.

Here is my array example. Have 10 more

[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]

Answer 1

Did you checked the Arrays.hashcode in Java 7 ?

 /**
 * 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;
}

I dont understand why @Marco13 mentioned " this is not returning unquie for arrays".

UPDATE

See @Macro13 comment for the reason why it cannot be unquie..

---

UPDATE

If we draw a graph using your input points, ( 18 elements) has one spike and 3 low values and the pattern goes.. if that is true.. you can find the mean of your Peak ( 1, 4, 8,12,16 ) and find the low Mean from remaining values.

So that you will be having Peak mean and Low mean . and you find the unquie number to represent these two also preserve the values using bijective algorithm described in here

This Alogirthm also provides formulas to reverse i.e take the Peak and Low mean from the unquie value.

To find unique pair < x; y >= x + (y + ( (( x +1 ) /2) * (( x +1 ) /2) ) )

Also refer Exercise 1 in pdf page 2 to reverse x and y.

For finding Mean and find paring value.

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);
}

for test

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));
}

You can use this the peak and low values to find the similar images.