Java 8 嵌套循环与流 &表现 [英] Java 8 nested loops with streams & performance

问题描述

In order to practise the Java 8 streams I tried converting the following nested loop to the Java 8 stream API. It calculates the largest digit sum of a^b (a,b < 100) and takes ~0.135s on my Core i5 760.

public static int digitSum(BigInteger x)
{
    int sum = 0;
    for(char c: x.toString().toCharArray()) {sum+=Integer.valueOf(c+"");}
    return sum;
}

@Test public void solve()
    {
        int max = 0;
        for(int i=1;i<100;i++)
            for(int j=1;j<100;j++)
                max = Math.max(max,digitSum(BigInteger.valueOf(i).pow(j)));
        System.out.println(max);
    }

My solution, which I expected to be faster because of the paralellism actually took 0.25s (0.19s without the parallel()):

int max =   IntStream.range(1,100).parallel()
            .map(i -> IntStream.range(1, 100)
            .map(j->digitSum(BigInteger.valueOf(i).pow(j)))
            .max().getAsInt()).max().getAsInt();

My questions

• did I do the conversion right or is there a better way to convert nested loops to stream calculations?
• why is the stream variant so much slower than the old one?
• why did the parallel() statement actually increased the time from 0.19s to 0.25s?

I know that microbenchmarks are fragile and parallelism is only worth it for big problems but for a CPU, even 0.1s is an eternity, right?

Update

I measure with the Junit 4 framework in Eclipse Kepler (it shows the time taken for executing a test).

My results for a,b<1000 instead of 100:

• traditional loop 186s
• sequential stream 193s
• parallel stream 55s

Update 2 Replacing sum+=Integer.valueOf(c+""); with sum+= c - '0'; (thanks Peter!) shaved off 10 whole seconds of the parallel method, bringing it to 45s. Didn't expect such a big performance impact!

Also, reducing the parallelism to the number of CPU cores (4 in my case) didn't do much as it reduced the time only to 44.8s (yes, it adds a and b=0 but I think this won't impact the performance much):

int max = IntStream.range(0, 3).parallel().
          .map(m -> IntStream.range(0,250)
          .map(i -> IntStream.range(1, 1000)
          .map(j->.digitSum(BigInteger.valueOf(250*m+i).pow(j)))
          .max().getAsInt()).max().getAsInt()).max().getAsInt();

解决方案

I have created a quick and dirty micro benchmark based on your code. The results are:

loop: 3192
lambda: 3140
lambda parallel: 868

So the loop and lambda are equivalent and the parallel stream significantly improves the performance. I suspect your results are unreliable due to your benchmarking methodology.

public static void main(String[] args) {
    int sum = 0;

    //warmup
    for (int i = 0; i < 100; i++) {
        solve();
        solveLambda();
        solveLambdaParallel();
    }

    {
        long start = System.nanoTime();
        for (int i = 0; i < 100; i++) {
            sum += solve();
        }
        long end = System.nanoTime();
        System.out.println("loop: " + (end - start) / 1_000_000);
    }
    {
        long start = System.nanoTime();
        for (int i = 0; i < 100; i++) {
            sum += solveLambda();
        }
        long end = System.nanoTime();
        System.out.println("lambda: " + (end - start) / 1_000_000);
    }
    {
        long start = System.nanoTime();
        for (int i = 0; i < 100; i++) {
            sum += solveLambdaParallel();
        }
        long end = System.nanoTime();
        System.out.println("lambda parallel : " + (end - start) / 1_000_000);
    }
    System.out.println(sum);
}

public static int digitSum(BigInteger x) {
    int sum = 0;
    for (char c : x.toString().toCharArray()) {
        sum += Integer.valueOf(c + "");
    }
    return sum;
}

public static int solve() {
    int max = 0;
    for (int i = 1; i < 100; i++) {
        for (int j = 1; j < 100; j++) {
            max = Math.max(max, digitSum(BigInteger.valueOf(i).pow(j)));
        }
    }
    return max;
}

public static int solveLambda() {
    return  IntStream.range(1, 100)
            .map(i -> IntStream.range(1, 100).map(j -> digitSum(BigInteger.valueOf(i).pow(j))).max().getAsInt())
            .max().getAsInt();
}

public static int solveLambdaParallel() {
    return  IntStream.range(1, 100)
            .parallel()
            .map(i -> IntStream.range(1, 100).map(j -> digitSum(BigInteger.valueOf(i).pow(j))).max().getAsInt())
            .max().getAsInt();
}

---

I have also run it with jmh which is more reliable than manual tests. The results are consistent with above (micro seconds per call):

Benchmark                                Mode   Mean        Units
c.a.p.SO21968918.solve                   avgt   32367.592   us/op
c.a.p.SO21968918.solveLambda             avgt   31423.123   us/op
c.a.p.SO21968918.solveLambdaParallel     avgt   8125.600    us/op

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