假设我有一些Java实体的列表,如
List<Entity> entities = Arrays.asList(entity1, entity2, entity3...);
我想将它减少为链式对象的一个实例,如:
class EntityChain {
private final Entity entity;
private final Optional<EntityChain> fallback;
private EntityChain(Builder builder) {
this.entity = builder.entity;
this.fallback = builder.fallback;
}
public static Builder builder() {
return new Builder();
}
public static final class Builder {
private Entity entity;
private Optional<EntityChain> fallback = Optional.empty();
public Builder withEntity(Entity entity) {
this.entity = entity;
return this;
}
public Builder withFallback(EntityChain fallback) {
this.fallback = Optional.of(fallback);
return this;
}
public EntityChain build() {
return new EntityChain(this);
}
}
}
EntityChain
是不可变的,并且有一个建设者
因此,结果将是EntityChain
实例,如:
chain
-> entity = entity1
-> fallback
-> entity = entity2
-> fallback
-> entity = entity3
-> fallback
...
是否可以通过一些神奇的Java 8流畅减少来实现这一目标? 是
Stream.reduce(U identity,
BiFunction<U, ? super T, U> accumulator,
BinaryOperator<U> combiner)
适用于此?以某种方式使用它的构建者?
答案 0 :(得分:2)
Supplier<EntityChain>
。算法构建实体链逆转:首先构建实体(n),然后构建实体(n-1),...实体(0)。
BiFunction<EntityChain, Entity, EntityChain> reducing =
(next, entity) -> Optional.ofNullable(next)
// create a builder with fallback if EntityChain present
.map(fallback -> EntityChain.builder().withFallback(fallback))
// create a builder without fallback
.orElseGet(EntityChain::builder)
//build the EntityChain
.withEntity(entity).build();
// combiner never be used in sequentially stream
BinaryOperator<EntityChain> rejectedInParallelStream = (t1, t2) -> {
//when you use parallel the chain order maybe changed, and the result is wrong.
throw new IllegalStateException("Can't be used in parallel stream!");
};
EntityChain chain = reverse(entities).
stream().reduce(null, reducing, rejectedInParallelStream);
//copy & reverse the copied List
static <T> List<T> reverse(List<T> list) {
List<T> it = list.stream().collect(Collectors.toList());
Collections.reverse(it);
return it;
}
-> entity = entity1
-> fallback
-> entity = entity2
-> fallback (empty)
答案 1 :(得分:1)
您可以在需要时通过延迟构建EntityChain
来实现。我使用Supplier<EntityChain>
完成了它。
BiFunction<Supplier<EntityChain>, Entity, Supplier<EntityChain>> reducing =
(initializer, entity) ->
// get the EntityChain instance when get() called.
() -> Optional.ofNullable(initializer.get())
// create a builder with fallback if EntityChain present
.map(fallback -> EntityChain.builder().withFallback(fallback))
// create a builder without fallback
.orElseGet(EntityChain::builder)
//build the EntityChain
.withEntity(entity).build();
// combiner never be used in sequentially stream
BinaryOperator<Supplier<EntityChain>> rejectedInParallelStream = (t1, t2) -> {
throw new IllegalStateException("Can't be used in parallel stream!");
};
EntityChain chain = reverse(entities).
stream().reduce(() -> null, reducing, rejectedInParallelStream)
//when the initializer chain built/reduced,
//calling the get() to fetch EntityChain header
.get();
//copy & reverse the copied List
static <T> List<T> reverse(List<T> list) {
List<T> it = list.stream().collect(Collectors.toList());
Collections.reverse(it);
return it;
}
-> entity = entity1
-> fallback
-> entity = entity2
-> fallback (empty)
答案 2 :(得分:0)
为了将列表缩减为构建器,您需要一个累加器(Builder::withEntity
)和一个组合器(Builder::combine
):
public class ChainBuilder {
public static class Entity {
int data;
public Entity(int i) {
data = i;
}
@Override
public String toString() {
return "Entity [data=" + data + "]";
}
}
public static class EntityChain {
private Entity entity;
private Optional<EntityChain> fallback = null;
@Override
public String toString() {
return "EntityChain [entity=" + entity + ", fallback=" + fallback + "]";
}
public static class Builder {
private EntityChain chain = null;
public Builder() { }
private static EntityChain newChainLink(Entity e){
EntityChain n = new EntityChain();
n.entity = e;
n.fallback = Optional.empty();
return n;
}
/** accumulator, attaches to the end of the chain */
public Builder withEntity(Entity e) {
if(chain == null) {
chain = newChainLink(e);
} else {
EntityChain last = getLast();
last.fallback = Optional.of(newChainLink(e));
}
return this;
}
/** combiner, glues two chains together */
public Builder combine(Builder u) {
if(u.chain != null) {
getLast().fallback = Optional.of(u.chain);
}
return this;
}
/** returns the end of the chain */
private EntityChain getLast() {
EntityChain link = chain;
while(link.fallback.isPresent()){
link = link.fallback.get();
}
return link;
}
public EntityChain build() {
return chain;
}
}
}
public static void main(String[] args) {
List<Entity> entities = Arrays.asList(new Entity(1), new Entity(2), new Entity(3));
final Builder reduced = entities.stream().reduce(new EntityChain.Builder(),
(t,u)->t.withEntity(u),
(t,u)->t.combine(u));
System.out.println(reduced.build());
}
}
打印:
EntityChain [entity = Entity [data = 1],fallback = Optional [EntityChain [entity = Entity [data = 2],fallback = Optional [EntityChain [entity = Entity [data = 3],fallback = Optional.empty] ]]]]