在Java lambda中使用两个流来计算协方差

Question

假设我有两个double数组.我一直在尝试使用Java 8中的Stream.我想我已经理解了主要的想法,但后来我意识到我不确定如何同时操作两个Streams.

例如,我想计算两个数组的协方差.

public class foo {

    public static double mean(double[] xs) {
        return Arrays.stream(xs).average().getAsDouble();
}

    public static void main(String[] args) {
        double[] xs = {1, 2, 3, 4, 5, 6, 7, 8, 9};
        double[] ys = {1517.93, 1757.78, 1981.1, 2215.73, 2942.66, 3558.32, 4063.91, 4521.16, 5101.76, 5234.12};

        System.out.println("Mean of xs: " + mean(xs));
        double xs_sumDeviation = Arrays.stream(xs)
            .boxed()
            .mapToDouble(d -> d.doubleValue() - mean(xs))
            .sum();
       // Covariance
        double covXY = Arrays.stream(xs, ys)
            .mapToDouble(x,y -> {
                  double numerator = (x-mean(xs)* (y-mean(ys);
                  double denominator = Math.sqrt((x-mean(xs)* (x-mean(xs));
                  return numerator / denominator;
             })
            .sum();

    }
}

谢谢你的建议.

尝试1.

public static double covariance(double[] xs, double[] ys) {
    double xmean = mean(xs);
    double ymean = mean(ys);
    double numerator = IntStream.range(0, Math.min(xs.length, ys.length))
            .parallel()
            .mapToDouble(i -> (xs[i] - xmean) * (ys[i] - ymean))
            .sum();
    double denominator = Math.sqrt(IntStream.range(0, xs.length)
            .parallel()
            .mapToDouble(i -> (xs[i] - xmean) * (xs[i] - xmean))
            .sum());
    return numerator / denominator;

Answer 1

在其他编程语言中,存在某种类型的zip函数,它需要几个迭代,并返回一个迭代器,它聚合来自每个迭代的元素.例如,请参阅Python库中的函数zip.

尽管可以在Java中创建类似的函数,但很难以这种方式实现它,它支持高效的并行执行.但是,Java中有一种常用的模式,有点不同.在您的情况下,它可能看起来如下:

public static double covariance(double[] xs, double[] ys) {
    double xmean = mean(xs);
    double ymean = mean(ys);
    return IntStream.range(0, Math.min(xs.length, ys.length))
        .parallel()
        .mapToDouble(i -> {
                double numerator = (xs[i] - xmean) * (ys[i] - ymean);
                double denominator = ...;
                return numerator / denominator;
            })
        .sum();
}

您可以IntStream使用索引创建包含所有索引的元素,而不是组合两个流,而是访问不同集合的元素.只要集合支持随机访问操作,这种方法就能很好地工作.