是否可以使用Java Stream API查找数字集合的中位数?我有一个List<Double>
未排序的类型的变量。
从JDK9到14的javadocs似乎表明这median()
不是的有效方法DoubleStream
。平均值,最小值,最大值,计数和总和是有效的聚合函数。
这median()
不是有效方法,因为它需要在流传输之前进行排序的Collection?
这适用于average()
:
// Function to find average element from a List of Integers in Java 9 and above
public static Double getAverageWithStream(List<Double> list) {
OptionalDouble average = list.stream() // Stream<Double>
.mapToDouble(v -> v) // DoubleStream
.average(); // OptionalDouble
// Print out a message about the 'average' variable's value
// Note: ifPresentOrElse() was introduced in JDK9
average.ifPresentOrElse(
// message the value if one exists
(value) -> {
System.out.println("The average value is " + value);
},
// Alert the user that there is no value
() -> {
System.out.println("No average could be determined!");
}
); // end ifPresentOrElse()
return average.orElse(Double.NaN);
} // end getAverageWithStream()
根据@Robert Bain留下的评论,我的方法需要按以下方式进行更改以支持中位数聚合:
/**
* Function to find median element from a List of Integers in Java 9 and above
* @param list List<Double>
* @return Double
*/
public static Double getMedianWithStream(List<Double> list) {
DoubleStream sortedNumbers = list.stream() // Stream<Double>
.mapToDouble(v -> v) // DoubleStream
.sorted(); // DoubleStream sorted
OptionalDouble median = (
list.size() % 2 == 0 ?
sortedNumbers.skip((list.size() / 2) - 1)
.limit(2)
.average() :
sortedNumbers.skip(list.size() / 2)
.findFirst()
);
// Print out a message about the 'average' variable's value
// Note: ifPresentOrElse() was introduced in JDK9
median.ifPresentOrElse(
// message the value if one exists
(value) -> {
System.out.println("The median value is " + value);
},
// Alert the user that there is no value
() -> {
System.out.println("No median could be determined!");
}
); // end ifPresentOrElse()
return median.orElse(Double.NaN);
} // end getAverageWithStream()
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