有人可以帮我优化下面的代码吗?我不想在同一个列表上直播 3 次。我必须迭代同一个列表并应用不同的映射函数。 有人可以建议更好的解决方案吗-
List<Dummy> dummy = getDummyData(); //Assume we are getting data from some source
List<NewDummy> newDummyList = dummy.stream().map(eachDummy -> mapper.map(eachDummy, 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);
}
最佳答案
这似乎是自定义收集器的理想用例。但在此之前,我认为您可以将金额总和简化如下:
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的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);
}
现在我们准备好了:
dummy.stream().collect(toDummyObject(
eachDummy -> mapper.map(eachDummy, NewDummy.class),
someCondition));
关于lambda - Java8 Stream.map 在同一流上输出不同的映射函数?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52210527/