第六章：并行数据处理与性能

 并行流

　　假如你需要写一个方法，接受数字 n 作为参数，并返回从1到给定参数的所有数字的和。

public class DemoTest {

    public static void main(String[] args) {
        
        // 测试性能
        long t1 = System.currentTimeMillis();
        test(DemoTest::forSum, 1000000);
        System.out.println("forSum: " + " -> " + (System.currentTimeMillis() - t1) + " ms");
        t1 = System.currentTimeMillis();
        test(DemoTest::streamSum, 1000000);
        System.out.println("streamSum: " + " -> " + (System.currentTimeMillis() - t1) + " ms");
        t1 = System.currentTimeMillis();
        test(DemoTest::paralleSum, 1000000);
        System.out.println("paralleSum: " + " -> " + (System.currentTimeMillis() - t1) + " ms");
        t1 = System.currentTimeMillis();
        test(DemoTest::rangedSum, 1000000);
        System.out.println("rangedSum: " + " -> " + (System.currentTimeMillis() - t1) + " ms");
        t1 = System.currentTimeMillis();
        test(DemoTest::paralleRangedSum, 1000000);
        System.out.println("paralleRangedSum: " + " -> " + (System.currentTimeMillis() - t1) + " ms");
        
    }
    
    
    // stream求和
    public static long streamSum(long n) {
        return Stream.iterate(1L, i -> i+1)
                     .limit(n)
                     .reduce(0L, Long::sum);
    }
    // for求和
    public static long forSum(long n) {
        long result = 0;
        for (int i = 0; i <= n; i++) {
            result += i;
        }
        return result;
    }
    //paralle求和
     public static long paralleSum(long n) {
         return Stream.iterate(1L, i -> i+1)
                 .limit(n)
                 .parallel()// 将流转换为并行流
                 .reduce(0L, Long::sum);
     }
     
     // 测试性能--运行10次
     public static void test(Function<Long, Long> function, long n) {
         long result = 0;
         for (int i = 0; i < 10; i++) {
             result = function.apply(n);
         }
         System.out.println(result);
     }
     
     // 优化
     public static long rangedSum(long n) {
         return LongStream.rangeClosed(0L, n)
                           .reduce(0L, Long::sum);
     }
     public static long paralleRangedSum(long n) {
         return LongStream.rangeClosed(0L, n)
                           .parallel()
                           .reduce(0L, Long::sum);
     }
     
}

 

 总结：

源	可分解性
ArrayList	极佳
LinkedList	差
IntStream.range	极佳
Stream.iterate	差
HashSet	好
TreeSet	好

 

 

 

 

 

 

 

备注：

摘自文献：《Java8实战》（中文版）《Java8 in Action》（英文版）

 

代码（GitHub地址）: https://github.com/changlezhong/java8InAction