了解Java Stream reduce-并行与串行

我有Java代码，它使用lambda和streams来查找整数列表的平均值。它使用Stream.reduce（…）查找整数之和，然后计算平均值。即使删除parallel（）调用，代码也能正常工作。有人能解释一下原因吗？代码后面还有更多问题

import java.util.Arrays;
import java.util.List;

public class Temp {
    public static void main(String [] args){
        //For example, consider a list of a *series* of numbers in increasing order.
        List<Integer> numbers = Arrays.asList(1, 2, 3, 4, 5, 6);
        int n = numbers.get(numbers.size()-1);//n = 6 is the number of numbers in list.

        double expectedSum = (n * (n + 1))/2;//By mathematical formula for increasing series.

        Double sum = numbers
                //Take a stream of integers.
                .stream()
                //Split the stream into mini streams and calculate sum of each of mini stream.
                .parallel()
                //Reduce all the numbers in a stream to their sum.
                .reduce(
                        //start with a partial sum of 0.
                0.0,
                //For a stream, calculate the sum of all the numbers.
                (partialSum, nextNumber) -> partialSum + (double) nextNumber,
                //Add the sums of each mini stream.
                (onePartialSum, anotherPartialSum) -> (onePartialSum + anotherPartialSum)
        );

        System.out.println("Sum : expected value = " + expectedSum + ", actual value = " + sum);

        double expectedAverage = expectedSum/numbers.size();
        double average = sum/numbers.size();

        System.out.println("Average : expected value = " + expectedAverage + ", actual value = " + average);

    }
}

考虑下面的代码，该函数是一个

二进制运算符组合器

：

(onePartialSum, anotherPartialSum) -> (onePartialSum + anotherPartialSum)

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这条线实际上是如何并行和非并行工作的？我的理解是，在并行过程中，它取每个迷你流的总和，并将其添加到“总和”中。对吗？当流不平行时，它会做零+单个流的总和吗？如果是，则前一行代码的更准确替换将是

（sum，onePartialSum）->（sum+onePartialSum）

，其中sum已被初始化为零？？

并行：

Sum : expected value = 21.0, actual value = 21.0
Average : expected value = 3.5, actual value = 3.5

System.out.println(Arrays.asList(1, 2, 3, 4, 5).stream().parallel()
    .reduce(/*identity*/1, /*accumulator*/(s, s2) -> s + s2, /*combiner*/(s, s2) -> s + s2));

System.out.println(Arrays.asList(1, 2, 3, 4, 5).stream()
    .reduce(/*identity*/1, /*accumulator*/(s, s2) -> s + s2, /*combiner*/(s, s2) -> s + s2));

产出：20

累加器的参数是标识和每个流值

工作：

1+1, 1+2, 1+3, 1+4, 1+5 // Output of accumulator is 2,3,4,5,6

2+3 in one thread, 4+5 in another thread and then 5/*(2+3)*/+9/*(4+5)*/ and then 14/*(5+9)*/+6 = 20

1(identity) + 1(first element) = 2(new identity)
2(new identity) + 2(second element) = 4(new identity)
4(new identity) + 3(third element) = 7(new identity)
7(new identity) + 4(third element) = 11(new identity)
11(new identity) + 5(third element) = 16(new identity)

组合器的参数是累加器的输出

工作：

1+1, 1+2, 1+3, 1+4, 1+5 // Output of accumulator is 2,3,4,5,6

2+3 in one thread, 4+5 in another thread and then 5/*(2+3)*/+9/*(4+5)*/ and then 14/*(5+9)*/+6 = 20

1(identity) + 1(first element) = 2(new identity)
2(new identity) + 2(second element) = 4(new identity)
4(new identity) + 3(third element) = 7(new identity)
7(new identity) + 4(third element) = 11(new identity)
11(new identity) + 5(third element) = 16(new identity)

串联：

Sum : expected value = 21.0, actual value = 21.0
Average : expected value = 3.5, actual value = 3.5

System.out.println(Arrays.asList(1, 2, 3, 4, 5).stream().parallel()
    .reduce(/*identity*/1, /*accumulator*/(s, s2) -> s + s2, /*combiner*/(s, s2) -> s + s2));

System.out.println(Arrays.asList(1, 2, 3, 4, 5).stream()
    .reduce(/*identity*/1, /*accumulator*/(s, s2) -> s + s2, /*combiner*/(s, s2) -> s + s2));

产出：16

在串联模式下，不会执行组合器，累加器的参数为identity和每个流值，但每个流的结果设置为identity

工作：

1+1, 1+2, 1+3, 1+4, 1+5 // Output of accumulator is 2,3,4,5,6

2+3 in one thread, 4+5 in another thread and then 5/*(2+3)*/+9/*(4+5)*/ and then 14/*(5+9)*/+6 = 20

1(identity) + 1(first element) = 2(new identity)
2(new identity) + 2(second element) = 4(new identity)
4(new identity) + 3(third element) = 7(new identity)
7(new identity) + 4(third element) = 11(new identity)
11(new identity) + 5(third element) = 16(new identity)

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我希望您能理解。

“即使我删除了parallel（）调用，代码也能正常工作”——为什么它不能正常工作？您的累加器和合并器都正确实现，这意味着流将按照设计并行或顺序工作。@Slaw-我想了解合并器是如何并行或串行工作的。这就是我的最后一个问题asksIIRC，对于顺序流不会调用合并器，因为从来没有两组数据可以从多个线程合并。当并行时，流可以拆分源

拆分器

并使用多个线程；在这种情况下，每个线程都有自己的一组数据，需要在最后进行组合，因此需要组合器。