基于Map.get nullPointer优化的Java 8映射

Question

public class StartObject{
     private Something something;
     private Set<ObjectThatMatters> objectThatMattersSet;
}

public class Something{
     private Set<SomeObject> someObjecSet;
}

public class SomeObject {
     private AnotherObject anotherObjectSet;
}

public class AnotherObject{
     private Set<ObjectThatMatters> objectThatMattersSet;
}

public class ObjectThatMatters{
     private Long id;
}


private void someMethod(StartObject startObject) {
    Map<Long, ObjectThatMatters> objectThatMattersMap = StartObject.getSomething()
            .getSomeObject.stream()
            .map(getSomeObject::getAnotherObject)
            .flatMap(anotherObject-> anotherObject.getObjectThatMattersSet().stream())
            .collect(Collectors.toMap(ObjectThatMatters -> ObjectThatMatters.getId(), Function.identity()));
   Set<ObjectThatMatters > dbObjectThatMatters = new HashSet<>();
   try {
        dbObjectThatMatters.addAll( tartObject.getObjectThatMatters().stream().map(objectThatMatters-> objectThatMattersMap .get(objectThatMatters.getId())).collect(Collectors.toSet()));
    } catch (NullPointerException e) {
        throw new someCustomException();
    }
 startObject.setObjectThatMattersSet(dbObjectThatMatters);

给一个包含一组ObjectThatMatters的StartObject

和--包含已被所有有效ObjectThatMatters填充的数据库结构的东西。

当时，我希望将ObjectThatMatters的StartObject集交换为仅存在于某物范围内的有效的相应db对象

然后我比较了StartObject上的ObjectThatMatters集

和将它们替换为对象内的有效ObjectThatMatters。

和如果某些ObjectThatMatters没有有效的ObjectThatMatters，则抛出一个someCustomException

这个someMethod看起来很可怕，我怎样才能使它更易读？

已经尝试过将that更改为一个可选的，但这实际上没有帮助。

由于性能原因，使用Map而不是使用List.contains的列表，这是个好主意吗？ObjectThatMatters的总数通常是500个。

我不允许更改其他类的结构，我只向您展示影响此方法的字段，而不是每个字段，因为它们是非常丰富的对象。

Answer 1

您根本不需要一个映射步骤。第一个操作产生一个Map，首先可以用于生成所需的Set。由于可能有比您感兴趣的对象更多的对象，所以可以执行筛选操作。

因此，首先，将所需对象的ID收集到一个集合中，然后收集相应的db对象，由ID的Set进行过滤。您可以通过比较结果的Set大小和ID Set的大小来验证是否找到了所有的ID。

private void someMethod(StartObject startObject) {
    Set<Long> id = startObject.getObjectThatMatters().stream()
        .map(ObjectThatMatters::getId).collect(Collectors.toSet());

    HashSet<ObjectThatMatters> objectThatMattersSet =
        startObject.getSomething().getSomeObject().stream()
            .flatMap(so -> so.getAnotherObject().getObjectThatMattersSet().stream())
            .filter(obj -> id.contains(obj.getId()))
            .collect(Collectors.toCollection(HashSet::new));

    if(objectThatMattersSet.size() != id.size())
        throw new SomeCustomException();

    startObject.setObjectThatMattersSet(objectThatMattersSet);
}

这段代码生成一个HashSet；如果这不是一个要求，您可以只使用Collectors.toSet()来获得一个任意的Set实现。

甚至很容易发现哪些It丢失了：

private void someMethod(StartObject startObject) {
    Set<Long> id = startObject.getObjectThatMatters().stream()
        .map(ObjectThatMatters::getId)
        .collect(Collectors.toCollection(HashSet::new));// ensure mutable Set

    HashSet<ObjectThatMatters> objectThatMattersSet =
        startObject.getSomething().getSomeObject().stream()
            .flatMap(so -> so.getAnotherObject().getObjectThatMattersSet().stream())
            .filter(obj -> id.contains(obj.getId()))
            .collect(Collectors.toCollection(HashSet::new));

    if(objectThatMattersSet.size() != id.size()) {
        objectThatMattersSet.stream().map(ObjectThatMatters::getId).forEach(id::remove);
        throw new SomeCustomException("The following IDs were not found: "+id);
    }

    startObject.setObjectThatMattersSet(objectThatMattersSet);
}