Java 8 Streams:多个筛选器与复杂条件
Java 8 Streams:多个筛选器与复杂条件
提问于 2014-06-05 16:01:18
有时，您希望过滤具有多个条件的Stream：
myList.stream().filter(x -> x.size() > 10).filter(x -> x.isCool()) ...或者，您可以用一个复杂的条件和一个filter来做同样的事情
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，但我不知道。
第一种方法在可读性上胜出，但是哪种方法对性能更好呢？
回答 3
Stack Overflow用户
回答已采纳
发布于 2014-06-05 16:21:00
对于这两种选择，必须执行的代码非常相似，以至于您无法可靠地预测结果。底层对象结构可能会有所不同，但这对hotspot优化器并不构成挑战。因此，这取决于周围的其他条件，如果有任何不同的话，这些条件将导致更快的执行。
组合两个filter实例会创建更多的对象，因此也会产生更多的委托代码，但是如果你使用方法引用而不是lambda表达式，这种情况可能会改变，例如用filter(ItemType::isCool)替换filter(x -> x.isCool())。这样就消除了为lambda表达式创建的合成委托方法。因此，使用两个方法引用组合两个筛选器可能会创建与使用带有&&的lambda表达式的单个filter调用相同或更少的委托代码。
但是，如前所述，这种开销将被HotSpot优化器消除，可以忽略不计。
从理论上讲，两个过滤器可以比一个过滤器更容易并行化，但这只与相当计算密集的任务相关。
所以没有简单的答案。
底线是，不要认为这样的性能差异低于气味检测阈值。使用可读性更好的文档。
²…并且将需要执行后续阶段的并行处理的实现，这是标准流实现当前未采用的道路
Stack Overflow用户
发布于 2016-04-10 04:05:59
这个测试表明，您的第二个选项可以更好地执行。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}现在代码如下：
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());
}Stack Overflow用户
发布于 2018-11-14 04:54:56
这是@Hank D共享的6种不同的样本测试组合的结果。很明显，形式为u -> exp1 && exp2的谓词在所有情况下都具有很高的性能。
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}页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/24054773
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