Question

有时您想要使用多个条件过滤Stream：

myList.stream().filter(x -> x.size() > 10).filter(x -> x.isCool()) ...

或者您可以对复杂条件和单 filter执行相同操作：

myList.stream().filter(x -> x.size() > 10 && x -> x.isCool()) ...

我的猜测是第二种方法具有更好的性能特征，但我知道它。

第一种方法在可读性方面取胜，但性能更好？

Answer 1

必须为两个备选方案执行的代码非常相似，以至于无法可靠地预测结果。底层对象结构可能不同，但对热点优化器没有任何挑战。因此，如果存在任何差异，它将取决于其他周围条件，这将导致更快的执行。

组合两个过滤器实例会创建更多的对象，因此会产生更多的委托代码，但是如果使用方法引用而不是lambda表达式，这可能会发生变化，例如：将filter(x -> x.isCool())替换为filter(ItemType::isCool)。这样就消除了为lambda表达式创建的合成委托方法。因此，使用带有filter的lambda表达式，使用两个方法引用组合两个过滤器可能会创建相同或更少的委托代码，而不是单个&&调用。

但是，如上所述，HotSpot优化器将消除这种开销，并且可以忽略不计。

理论上，两个滤波器可以比单个滤波器更容易并行化，但这只与计算密集型任务相关¹。

所以没有简单的答案。

最重要的是，不要考虑低于气味检测阈值的这种性能差异。使用更具可读性的内容。

¹...并且需要一个实现并行处理后续阶段，这是标准Stream实现当前没有采用的一条道路

Answer 2

此测试表明您的第二个选项可以表现得更好。首先是结果，然后是代码：

one filter with predicate of form u -> exp1 && exp2, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=4142, min=29, average=41.420000, max=82}
two filters with predicates of form u -> exp1, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=13315, min=117, average=133.150000, max=153}
one filter with predicate of form predOne.and(pred2), list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=10320, min=82, average=103.200000, max=127}

现在代码：

enum Gender {
    FEMALE,
    MALE
}

static class User {
    Gender gender;
    int age;

    public User(Gender gender, int age){
        this.gender = gender;
        this.age = age;
    }

    public Gender getGender() {
        return gender;
    }

    public void setGender(Gender gender) {
        this.gender = gender;
    }

    public int getAge() {
        return age;
    }

    public void setAge(int age) {
        this.age = age;
    }
}

static long test1(List<User> users){
    long time1 = System.currentTimeMillis();
    users.stream()
            .filter((u) -> u.getGender() == Gender.FEMALE && u.getAge() % 2 == 0)
            .allMatch(u -> true);                   // least overhead terminal function I can think of
    long time2 = System.currentTimeMillis();
    return time2 - time1;
}

static long test2(List<User> users){
    long time1 = System.currentTimeMillis();
    users.stream()
            .filter(u -> u.getGender() == Gender.FEMALE)
            .filter(u -> u.getAge() % 2 == 0)
            .allMatch(u -> true);                   // least overhead terminal function I can think of
    long time2 = System.currentTimeMillis();
    return time2 - time1;
}

static long test3(List<User> users){
    long time1 = System.currentTimeMillis();
    users.stream()
            .filter(((Predicate<User>) u -> u.getGender() == Gender.FEMALE).and(u -> u.getAge() % 2 == 0))
            .allMatch(u -> true);                   // least overhead terminal function I can think of
    long time2 = System.currentTimeMillis();
    return time2 - time1;
}

public static void main(String... args) {
    int size = 10000000;
    List<User> users =
    IntStream.range(0,size)
            .mapToObj(i -> i % 2 == 0 ? new User(Gender.MALE, i % 100) : new User(Gender.FEMALE, i % 100))
            .collect(Collectors.toCollection(()->new ArrayList<>(size)));
    repeat("one filter with predicate of form u -> exp1 && exp2", users, Temp::test1, 100);
    repeat("two filters with predicates of form u -> exp1", users, Temp::test2, 100);
    repeat("one filter with predicate of form predOne.and(pred2)", users, Temp::test3, 100);
}

private static void repeat(String name, List<User> users, ToLongFunction<List<User>> test, int iterations) {
    System.out.println(name + ", list size " + users.size() + ", averaged over " + iterations + " runs: " + IntStream.range(0, iterations)
            .mapToLong(i -> test.applyAsLong(users))
            .summaryStatistics());
}

Answer 3

从性能的角度来看，复杂的过滤条件会更好，但是最好的性能将显示使用标准type WiderWindow = Omit<Window, keyof IgnoredWindowProps>;进行循环的老式方式是最好的选择。小数组上的差异可能是10个元素的差异的2倍左右，大数组上的差异不是那么大。
您可以看看我的GitHub project，在该处我对多个数组迭代选项进行了性能测试

对于小型阵列10个元素的吞吐量ops / s：对于中等10,000个元素的吞吐率操作：对于大型阵列1,000,000个元素，吞吐量ops / s：

注意：测试在

上运行

8个CPU
1 GB RAM
操作系统版本：16.04.1 LTS（Xenial Xerus）
java版本：1.8.0_121
jvm：-XX：+ UseG1GC -server -Xmx1024m -Xms1024m

Answer 4

这是@Hank D共享的6种不同的样本测试结果显然，在所有情况下，u -> exp1 && exp2形式的谓词都是高效的。

one filter with predicate of form u -> exp1 && exp2, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=3372, min=31, average=33.720000, max=47}
two filters with predicates of form u -> exp1, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=9150, min=85, average=91.500000, max=118}
one filter with predicate of form predOne.and(pred2), list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=9046, min=81, average=90.460000, max=150}

one filter with predicate of form u -> exp1 && exp2, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=8336, min=77, average=83.360000, max=189}
one filter with predicate of form predOne.and(pred2), list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=9094, min=84, average=90.940000, max=176}
two filters with predicates of form u -> exp1, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=10501, min=99, average=105.010000, max=136}

two filters with predicates of form u -> exp1, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=11117, min=98, average=111.170000, max=238}
one filter with predicate of form u -> exp1 && exp2, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=8346, min=77, average=83.460000, max=113}
one filter with predicate of form predOne.and(pred2), list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=9089, min=81, average=90.890000, max=137}

two filters with predicates of form u -> exp1, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=10434, min=98, average=104.340000, max=132}
one filter with predicate of form predOne.and(pred2), list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=9113, min=81, average=91.130000, max=179}
one filter with predicate of form u -> exp1 && exp2, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=8258, min=77, average=82.580000, max=100}

one filter with predicate of form predOne.and(pred2), list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=9131, min=81, average=91.310000, max=139}
two filters with predicates of form u -> exp1, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=10265, min=97, average=102.650000, max=131}
one filter with predicate of form u -> exp1 && exp2, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=8442, min=77, average=84.420000, max=156}

one filter with predicate of form predOne.and(pred2), list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=8553, min=81, average=85.530000, max=125}
one filter with predicate of form u -> exp1 && exp2, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=8219, min=77, average=82.190000, max=142}
two filters with predicates of form u -> exp1, list size 10000000, averaged over 100 runs: LongSummaryStatistics{count=100, sum=10305, min=97, average=103.050000, max=132}

Java 8 Streams：多个过滤器与复杂条件

4 个答案: