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calculating gini index

2015228094

I want to what type of Gini coeffcient used in item distribution and what is the formula to calculate Gini coeffcient and Gini index

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Accepted answers

MartinLiebig

Hi,

the one used in Trees and Weight By Gini can be found here:

https://github.com/rapidminer/rapidminer-studio/blob/master/src/main/java/com/rapidminer/operator/learner/tree/criterions/GiniIndexCriterion.java

Best,

Martin

2015228094

Thank you

Is this formula used for clustring kmean evalution?

do you have the formula as mathematical equation?

MartinLiebig

Hi,

for clustering its: https://github.com/rapidminer/rapidminer-studio/blob/master/src/main/java/com/rapidminer/operator/validation/clustering/exampledistribution/GiniCoefficient.java

or in code

		double sum = 0;
		for (int i = 0; i < x.length; i++) {
			sum = sum + x[i];
		}

		double mean = sum / n;

		sum = 0;
		for (int i = 0; i < x.length; i++) {
			for (int j = 0; j < x.length; j++) {
				sum = sum + Math.abs(x[i] - x[j]);
			}
		}

		if (mean == 0) {
			return 0.0;
		}

		return 1 - (sum / (2 * mean * (n * ((double) n - 1))));

Best,

Martin

All comments

MartinLiebig

Hi,

the one used in Trees and Weight By Gini can be found here:

https://github.com/rapidminer/rapidminer-studio/blob/master/src/main/java/com/rapidminer/operator/learner/tree/criterions/GiniIndexCriterion.java

Best,

Martin

2015228094

Thank you

Is this formula used for clustring kmean evalution?

do you have the formula as mathematical equation?

MartinLiebig

Hi,

for clustering its: https://github.com/rapidminer/rapidminer-studio/blob/master/src/main/java/com/rapidminer/operator/validation/clustering/exampledistribution/GiniCoefficient.java

or in code

		double sum = 0;
		for (int i = 0; i < x.length; i++) {
			sum = sum + x[i];
		}

		double mean = sum / n;

		sum = 0;
		for (int i = 0; i < x.length; i++) {
			for (int j = 0; j < x.length; j++) {
				sum = sum + Math.abs(x[i] - x[j]);
			}
		}

		if (mean == 0) {
			return 0.0;
		}

		return 1 - (sum / (2 * mean * (n * ((double) n - 1))));

Best,

Martin