Difference between revisions of "User:Skilgannon/Free Code"
Jump to navigation
Jump to search
Skilgannon (talk | contribs) (did some more testing, and clarify a bit) |
Skilgannon (talk | contribs) m (prettify it with tex math =)) |
||
Line 1: | Line 1: | ||
− | This is a neat method that I made up. It takes array of 'indexes' (be it guess factors, or indexes where you are logging hits) between 0 and max and returns a double between 0 and 1 of how 'clustered' your array is. 1 if all the values are the same, and 0 if there are infinite values spread perfectly evenly. Note, this is very different from a standard deviation calculation. In this code there can be as many 'dense' points on the graph as you want, and it won't try to accommodate them all from one mean. Instead, it relies on the fact that (d+k)^2 | + | This is a neat method that I made up. It takes array of 'indexes' (be it guess factors, or indexes where you are logging hits) between 0 and max and returns a double between 0 and 1 of how 'clustered' your array is. 1 if all the values are the same, and 0 if there are infinite values spread perfectly evenly. Note, this is very different from a standard deviation calculation. In this code there can be as many 'dense' points on the graph as you want, and it won't try to accommodate them all from one mean. Instead, it relies on the fact that |
+ | <math> (d+k)^2 > d^2 > (d-k)^2 </math> for any <math> d > k > 0 </math>. | ||
<pre> | <pre> |
Revision as of 18:00, 13 April 2010
This is a neat method that I made up. It takes array of 'indexes' (be it guess factors, or indexes where you are logging hits) between 0 and max and returns a double between 0 and 1 of how 'clustered' your array is. 1 if all the values are the same, and 0 if there are infinite values spread perfectly evenly. Note, this is very different from a standard deviation calculation. In this code there can be as many 'dense' points on the graph as you want, and it won't try to accommodate them all from one mean. Instead, it relies on the fact that <math> (d+k)^2 > d^2 > (d-k)^2 </math> for any <math> d > k > 0 </math>.
public static double clustering(float[] indexes, float max){ float[] sorted = new float[indexes.length]; System.arraycopy(indexes,0,sorted,0,indexes.length); java.util.Arrays.sort(sorted); double clustering = sorted[0] + max - sorted[sorted.length - 1]; clustering *= clustering; for(int i = 1; i < sorted.length; i++){ double diff = sorted[i] - sorted[i-1]; clustering += diff*diff; } return clustering/(max*max); }