我一直在寻找一种方法来计算给定列表中每个值的百分等级,到目前为止我都没有成功。
org.apache.commons.math3
为您提供了一种从值列表中获取第p个百分位数的方法,但我想要的是相反的。我想对列表中的每个值进行排名。是否有人知道一个库或Apache公共数学方法来实现这一目标?
例如:给定一个值列表{1,2,3,4,5}
,我希望每个值都具有百分等级,最大百分位数为99或100,最小值为0或1。
更新的代码:
public class TestPercentile {
public static void main(String args[]) {
double x[] = { 10, 11, 12, 12, 12, 12, 15, 18, 19, 20 };
calculatePercentiles(x);
}
public static void calculatePercentiles(double[] arr) {
for (int i = 0; i < arr.length; i++) {
int count = 0;
int start = i;
if (i > 0) {
while (i > 0 && arr[i] == arr[i - 1]) {
count++;
i++;
}
}
double perc = ((start - 0) + (0.5 * count));
perc = perc / (arr.length - 1);
for (int k = 0; k < count + 1; k++)
System.out.println("Percentile for value " + (start + k + 1)
+ " = " + perc * 100);
}
}}
Sample Output:
Percentile for value 1 = 0.0
Percentile for value 2 = 11.11111111111111
Percentile for value 3 = 22.22222222222222
Percentile for value 4 = 50.0
Percentile for value 5 = 50.0
Percentile for value 6 = 50.0
Percentile for value 7 = 50.0
Percentile for value 8 = 77.77777777777779
Percentile for value 9 = 88.88888888888889
Percentile for value 10 = 100.0
有人可以告诉我这是否正确以及是否有一个可以更干净地执行此操作的库?
谢谢!
答案 0 :(得分:6)
这实际上取决于你对百分位数的定义。以下是使用NaturalRanking并重新缩放到0-1间隔的解决方案。很好的是,NaturalRanking有一些处理相同值和nans已经实现的策略。
import java.util.Arrays;
import org.apache.commons.math3.stat.ranking.NaNStrategy;
import org.apache.commons.math3.stat.ranking.NaturalRanking;
import org.apache.commons.math3.stat.ranking.TiesStrategy;
public class Main {
public static void main(String[] args) {
double[] arr = {Double.NaN, 10, 11, 12, 12, 12, 12, 15, 18, 19, 20};
PercentilesScaledRanking ranking = new PercentilesScaledRanking(NaNStrategy.REMOVED, TiesStrategy.MAXIMUM);
double[] ranks = ranking.rank(arr);
System.out.println(Arrays.toString(ranks));
//prints:
//[0.1, 0.2, 0.6, 0.6, 0.6, 0.6, 0.7, 0.8, 0.9, 1.0]
}
}
class PercentilesScaledRanking extends NaturalRanking {
public PercentilesScaledRanking(NaNStrategy nanStrategy, TiesStrategy tiesStrategy) {
super(nanStrategy, tiesStrategy);
}
@Override
public double[] rank(double[] data) {
double[] rank = super.rank(data);
for (int i = 0; i < rank.length; i++) {
rank[i] = rank[i] / rank.length;
}
return rank;
}
}