java – 计算列表中每个值的百分位数
作者:互联网
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,到目前为止我还没有成功.
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,是否有一个库可以更干净地做到这一点?
谢谢!
解决方法:
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.很自然,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;
}
}
标签:java,statistics,percentile,apache-commons-math 来源: https://codeday.me/bug/20190517/1120867.html