Lambda Java8 stream.map在同一个流上输出不同的映射函数?
有人能帮我优化下面的代码吗。我不想在同一个列表上流3次。我必须在同一个列表上迭代并应用不同的映射函数。 有人能提出更好的解决方案吗Lambda Java8 stream.map在同一个流上输出不同的映射函数?,lambda,java-8,functional-programming,java-stream,Lambda,Java 8,Functional Programming,Java Stream,有人能帮我优化下面的代码吗。我不想在同一个列表上流3次。我必须在同一个列表上迭代并应用不同的映射函数。 有人能提出更好的解决方案吗 List<Dummy> dummy = getDummyData(); //Assume we are getting data from some source List<NewDummy> newDummyList = dummy.stream().map(eachDummy -> mapper.map(eachDummy, New
List<Dummy> dummy = getDummyData(); //Assume we are getting data from some source
List<NewDummy> newDummyList = dummy.stream().map(eachDummy -> mapper.map(eachDummy, NewDummy.class)).collect(Collectors.toList());
if(someCondition) {
final BigDecimal amount1 = dummy.stream().filter(eachDummy -> Optional.ofNullable(eachDummy.getAmount1()).isPresent())
.map(Dummy::getAmount1).reduce(BigDecimal.ZERO, BigDecimal::add);
final BigDecimal amount2 = dummy.stream().filter(eachDummy -> Optional.ofNullable(eachDummy.getAmount2()).isPresent())
.map(Dummy::getAmount2).reduce(BigDecimal.ZERO, BigDecimal::add);
return new DummyObject(newDummyList, amount1, amount2);
} else {
return new DummyObject(newDummyList);
}
List dummy=getDummyData()//假设我们从某个来源获取数据
List newDummyList=dummy.stream().map(eachDummy->mapper.map(eachDummy,NewDummy.class)).collect(Collectors.toList());
如果(某些条件){
final BigDecimal amount1=dummy.stream().filter(eachDummy->Optional.ofNullable(eachDummy.getAmount1()).isPresent())
.map(Dummy::getAmount1).reduce(BigDecimal.ZERO,BigDecimal::add);
final BigDecimal amount2=dummy.stream().filter(eachDummy->Optional.ofNullable(eachDummy.getAmount2()).isPresent())
.map(Dummy::getAmount2).reduce(BigDecimal.ZERO,BigDecimal::add);
返回新的DummyObject(newDummyList,数量1,数量2);
}否则{
返回新的DummyObject(newDummyList);
}
这似乎是定制收集器的理想用例。但在此之前,我认为您可以将金额的总和简化如下:
BigDecimal amount1 = dummy.stream()
.map(Dummy::getAmount1)
.filter(Objects::nonNull)
.reduce(BigDecimal::add).orElse(BigDecimal.ZERO);
现在,自定义收集器。您可以将Dummy
的实例累积到静态实用程序方法内专用本地类的实例中:
static Collector<Dummy, ?, DummyObject> toDummyObject(
Function<Dummy, NewDummy> mapper,
boolean someCondition) {
class Accumulator {
List<NewDummy> newDummyList = new ArrayList<>();
BigDecimal amount1 = BigDecimal.ZERO;
BigDecimal amount2 = BigDecimal.ZERO;
public void add(Dummy dummy) {
newDummyList.add(mapper.apply(dummy));
}
public void addAndSum(Dummy dummy) {
if (dummy.getAmount1() != null) amount1 = amount1.add(dummy.getAmount1());
if (dummy.getAmount2() != null) amount2 = amount2.add(dummy.getAmount2());
add(dummy);
}
public Accumulator merge(Accumulator another) {
newDummyList.addAll(another.newDummyList);
return this;
}
public Accumulator mergeAndSum(Accumulator another) {
amount1 = amount1.add(another.amount1);
amount2 = amount2.add(another.amount2);
return merge(another);
}
public DummyObject finish() {
return someCondition ?
new DummyObject(newDummyList, amount1, amount2) :
new DummyObject(newDummyList);
}
}
return Collector.of(
Accumulator::new,
someCondition ? Accumulator::addAndSum : Accumulator::add,
someCondition ? Accumulator::mergeAndSum : Accumulator::merge,
Accumulator::finish);
}
我同意收集器似乎是这里最好的选择
但是,我不想实现一个非常专门的收集器
,我更喜欢只实现一些通用的“构建块”,然后使用这些块来组成我在给定情况下需要的收集器
假设:
interface Mapper<T> {
T map(Dummy dummy, Class<T> type);
}
下面是我如何编写特定于用例的收集器
s:
private static Collector<Dummy, ?, DummyObject> toDummyObjectWithoutSums(Mapper<NewDummy> mapper) {
return Collectors.collectingAndThen(toNewDummyList(mapper), DummyObject::new);
}
private static Collector<Dummy, ?, List<NewDummy>> toNewDummyList(Mapper<NewDummy> mapper) {
return Collectors.mapping(dummy -> mapper.map(dummy, NewDummy.class), Collectors.toList());
}
private static Collector<Dummy, ?, DummyObject> toDummyObjectWithSums(Mapper<NewDummy> mapper) {
return ExtraCollectors.collectingBoth(
toNewDummyList(mapper),
sumGroupCollector(),
(newDummyList, amountSumPair) -> new DummyObject(
newDummyList, amountSumPair.getAmountSum1(), amountSumPair.getAmountSum2()
)
);
}
private static Collector<Dummy, ?, AmountSumPair> sumGroupCollector() {
return ExtraCollectors.collectingBoth(
summingAmount(Dummy::getAmount1),
summingAmount(Dummy::getAmount2),
AmountSumPair::new
);
}
static Collector<Dummy, ?, BigDecimal> summingAmount(Function<Dummy, BigDecimal> getter) {
return Collectors.mapping(getter,
ExtraCollectors.filtering(Objects::nonNull,
ExtraCollectors.summingBigDecimal()
)
);
}
private static class AmountSumPair {
private final BigDecimal amountSum1;
private final BigDecimal amountSum2;
// constructor + getters
}
下面是BiCollectorHandler
类(由extracollector内部使用。同时收集两个):
final类BiCollectorHandler{
私人最终收藏家1;
私人最终收藏家2;
双收集器处理程序(收集器收集器1、收集器收集器2){
this.collector1=collector1;
this.collector2=collector2;
}
双累加器(双累加器){
返回新的BiAccumulator(collector1.supplier().get(),collector2.supplier().get());
}
末级双累加器{
最终A1 acc1;
最终A2 acc2;
专用双累加器(A1 acc1、A2 acc2){
this.acc1=acc1;
this.acc2=acc2;
}
无效接受(T){
收集器1.累加器().接受(acc1,t);
收集器2.累加器().接受(acc2,t);
}
双累加器联合收割机(双累加器其他){
A1 combined1=collector1.combiner().apply(acc1,other.acc1);
A2 combined2=collector2.combiner().apply(acc2,other.acc2);
返回新的双累加器(combined1,combined2);
}
R表面处理(双功能双整理器){
R1结果1=collector1.finisher().apply(acc1);
R2结果2=collector2.finisher().apply(acc2);
返回biFinisher.apply(结果1、结果2);
}
}
}
您的eachDummy::getAmount1
和eachDummy::getAmount2
应该分别是Dummy::getAmount1
和Dummy::getAmount1
。@TomaszLinkowski yes..这是正确的。。意志update@federico谢谢你提出了一个很好的选择。就性能而言,您不认为在原始方法中使用Reduce来计算和的性能会优于您在这里建议的代码吗。基本上,如果我想比较你建议的方法和讨论中的方法,我应该怎么做?@prats谢谢你的话。在这里搜索jmh microbenchmark,您会发现非常有用的信息。我相信(理论上)我的版本会比你的稍好一点,只是一点点。两个版本具有相同的时间复杂性。我认为有3个单独的循环或流是可以的,也许你可以重构一个私有方法。调试代码后,我意识到组合器(mergeAndSum或merge)将永远不会被调用,因为这是一个顺序流。因此,我们可以将其更改为并行流以利用combiners@prats您可以尝试并测量,但我怀疑您是否会发现这项工作有任何改进+1(我打赌它花费了相当长的时间编写),顺便说一句,BiCollector
可能会出现在jdk-12上。@Eugene Ha,在我之前;)很高兴知道,谢谢你的信息!我希望这个特性能够应用到JDK12上。是的,写起来花了一段时间——比我预期的要长一点,但这就是SO的问题:解决像这里这样的问题一旦开始就会上瘾;)Tomasz,这是一项伟大的工作,我同意拥有这些构建块是一种更好的方法。我也非常喜欢你最初的回答,我认为它显示了硬币的另一面,也就是说,你不需要专门的收款人,只要将金额的总和重构成一个私人方法,你就可以在列表上流3次。
@federicoperataschaffner谢谢!)关于我最初的回答:这实际上是错误的,因为我误解了这个问题,只流了一次List
,流了两次List
。当然,列表
的三重流也可能是一个选项,但由于OP写了“假设我们从某个源获取数据”,我认为这样的流被认为是昂贵的。更新:“BiCollector”功能作为收集器出现在JDK12上。teeing
:。
Collector<Dummy, ?, DummyObject> dummyObjectCollector = someCondition
? toDummyObjectWithSums(mapper)
: toDummyObjectWithoutSums(mapper);
return dummy.stream().collect(dummyObjectCollector);
private static Collector<Dummy, ?, DummyObject> toDummyObjectWithoutSums(Mapper<NewDummy> mapper) {
return Collectors.collectingAndThen(toNewDummyList(mapper), DummyObject::new);
}
private static Collector<Dummy, ?, List<NewDummy>> toNewDummyList(Mapper<NewDummy> mapper) {
return Collectors.mapping(dummy -> mapper.map(dummy, NewDummy.class), Collectors.toList());
}
private static Collector<Dummy, ?, DummyObject> toDummyObjectWithSums(Mapper<NewDummy> mapper) {
return ExtraCollectors.collectingBoth(
toNewDummyList(mapper),
sumGroupCollector(),
(newDummyList, amountSumPair) -> new DummyObject(
newDummyList, amountSumPair.getAmountSum1(), amountSumPair.getAmountSum2()
)
);
}
private static Collector<Dummy, ?, AmountSumPair> sumGroupCollector() {
return ExtraCollectors.collectingBoth(
summingAmount(Dummy::getAmount1),
summingAmount(Dummy::getAmount2),
AmountSumPair::new
);
}
static Collector<Dummy, ?, BigDecimal> summingAmount(Function<Dummy, BigDecimal> getter) {
return Collectors.mapping(getter,
ExtraCollectors.filtering(Objects::nonNull,
ExtraCollectors.summingBigDecimal()
)
);
}
private static class AmountSumPair {
private final BigDecimal amountSum1;
private final BigDecimal amountSum2;
// constructor + getters
}
final class ExtraCollectors {
static Collector<BigDecimal, ?, BigDecimal> summingBigDecimal() {
return Collectors.reducing(BigDecimal.ZERO, BigDecimal::add);
}
static <T, A, R> Collector<T, A, R> filtering(
Predicate<T> filter, Collector<T, A, R> downstream) {
return Collector.of(
downstream.supplier(),
(A acc, T t) -> {
if (filter.test(t)) {
downstream.accumulator().accept(acc, t);
}
},
downstream.combiner(),
downstream.finisher(),
downstream.characteristics().toArray(new Collector.Characteristics[0])
);
}
static <T, R1, R2, R> Collector<T, ?, R> collectingBoth(
Collector<T, ?, R1> collector1, Collector<T, ?, R2> collector2, BiFunction<R1, R2, R> biFinisher) {
return collectingBoth(new BiCollectorHandler<>(collector1, collector2), biFinisher);
}
// method needed to capture A1 and A2
private static <T, A1, R1, A2, R2, R> Collector<T, ?, R> collectingBoth(
BiCollectorHandler<T, A1, R1, A2, R2> biCollectorHandler, BiFunction<R1, R2, R> biFinisher) {
return Collector.<T, BiCollectorHandler<T, A1, R1, A2, R2>.BiAccumulator, R>of(
biCollectorHandler::newBiAccumulator,
BiCollectorHandler.BiAccumulator::accept,
BiCollectorHandler.BiAccumulator::combine,
biAccumulator -> biAccumulator.finish(biFinisher)
);
}
}
final class BiCollectorHandler<T, A1, R1, A2, R2> {
private final Collector<T, A1, R1> collector1;
private final Collector<T, A2, R2> collector2;
BiCollectorHandler(Collector<T, A1, R1> collector1, Collector<T, A2, R2> collector2) {
this.collector1 = collector1;
this.collector2 = collector2;
}
BiAccumulator newBiAccumulator() {
return new BiAccumulator(collector1.supplier().get(), collector2.supplier().get());
}
final class BiAccumulator {
final A1 acc1;
final A2 acc2;
private BiAccumulator(A1 acc1, A2 acc2) {
this.acc1 = acc1;
this.acc2 = acc2;
}
void accept(T t) {
collector1.accumulator().accept(acc1, t);
collector2.accumulator().accept(acc2, t);
}
BiAccumulator combine(BiAccumulator other) {
A1 combined1 = collector1.combiner().apply(acc1, other.acc1);
A2 combined2 = collector2.combiner().apply(acc2, other.acc2);
return new BiAccumulator(combined1, combined2);
}
<R> R finish(BiFunction<R1, R2, R> biFinisher) {
R1 result1 = collector1.finisher().apply(acc1);
R2 result2 = collector2.finisher().apply(acc2);
return biFinisher.apply(result1, result2);
}
}
}