如何组合两个对象元素
How to combine two object elements
我想将两个单独的结果合并为一个结果。
我的第一个结果是第一个算法的项目分数:
PredictedResult(List(ItemScore(140849,0.6259532295250041,0.0,0.0,0.0)),List())
我的第二个结果是第二个算法的规则分数:
PredictedResult(List(),List(Rule(Set(140855),List(ItemScore(368788,0.0,1.3516984090509725E-5,0.1111111111111111,38.59207094418362)))))
您可以看到第一个结果有一个空列表。此列表填充在第二个结果中。这也适用于项目分数,只是反过来。
服务 class 只接受两个结果之一,而不是合并。
Serving.scala:
package org.template
import org.apache.predictionio.controller.LServing
class Serving
extends LServing[Query, PredictedResult] {
override
def serve(query: Query,
predictedResults: Seq[PredictedResult]): PredictedResult = {
println(predictedResults(0))
println(predictedResults(1))
// Returning
predictedResults(0)
}
}
Engine.json:
package org.template
import org.apache.predictionio.controller.EngineFactory
import org.apache.predictionio.controller.Engine
// Query most similar (top num) items to the given
case class Query(items: Set[String], num: Int) extends Serializable
case class PredictedResult(itemScores: List[ItemScore], rules: List[Rule])
case class Rule(cond: Set[String], itemScores: List[ItemScore])
extends Serializable
case class ItemScore(item: String, score: Double, support: Double, confidence: Double, lift: Double) extends Serializable with
Ordered[ItemScore] {
def compare(that: ItemScore) = this.score.compare(that.score)
}
当我点击查询时,现在我只得到一个结果:
{"itemScores":[{"item":"140849","score":0.6259532295250041,"support":0.0,"confidence":0.0,"lift":0.0}],"rules":[]}
预期输出(算法 1 的 ItemScore 结合算法 2 的规则):
PredictedResult(List(ItemScore(140849,0.6259532295250041,0.0,0.0,0.0)),List(Rule(Set(140855),List(ItemScore(368788,0.0,1.3516984090509725E-5,0.1111111111111111,38.59207094418362)))))
像这样尝试列表连接++
case class ItemScore(i: Int)
case class Rule(s: String, itemScores: List[ItemScore])
case class PredictedResult(itemScores: List[ItemScore], rules: List[Rule])
val pr1 = PredictedResult(List(ItemScore(1)), Nil)
val pr2 = PredictedResult(Nil, List(Rule("rule1", List(ItemScore(2)))))
PredictedResult(pr1.itemScores ++ pr2.itemScores, pr1.rules ++ pr2.rules)
输出
res0: PredictedResult = PredictedResult(List(ItemScore(1)),List(Rule(rule1,List(ItemScore(2)))))
我想将两个单独的结果合并为一个结果。 我的第一个结果是第一个算法的项目分数:
PredictedResult(List(ItemScore(140849,0.6259532295250041,0.0,0.0,0.0)),List())
我的第二个结果是第二个算法的规则分数:
PredictedResult(List(),List(Rule(Set(140855),List(ItemScore(368788,0.0,1.3516984090509725E-5,0.1111111111111111,38.59207094418362)))))
您可以看到第一个结果有一个空列表。此列表填充在第二个结果中。这也适用于项目分数,只是反过来。
服务 class 只接受两个结果之一,而不是合并。
Serving.scala:
package org.template
import org.apache.predictionio.controller.LServing
class Serving
extends LServing[Query, PredictedResult] {
override
def serve(query: Query,
predictedResults: Seq[PredictedResult]): PredictedResult = {
println(predictedResults(0))
println(predictedResults(1))
// Returning
predictedResults(0)
}
}
Engine.json:
package org.template
import org.apache.predictionio.controller.EngineFactory
import org.apache.predictionio.controller.Engine
// Query most similar (top num) items to the given
case class Query(items: Set[String], num: Int) extends Serializable
case class PredictedResult(itemScores: List[ItemScore], rules: List[Rule])
case class Rule(cond: Set[String], itemScores: List[ItemScore])
extends Serializable
case class ItemScore(item: String, score: Double, support: Double, confidence: Double, lift: Double) extends Serializable with
Ordered[ItemScore] {
def compare(that: ItemScore) = this.score.compare(that.score)
}
当我点击查询时,现在我只得到一个结果:
{"itemScores":[{"item":"140849","score":0.6259532295250041,"support":0.0,"confidence":0.0,"lift":0.0}],"rules":[]}
预期输出(算法 1 的 ItemScore 结合算法 2 的规则):
PredictedResult(List(ItemScore(140849,0.6259532295250041,0.0,0.0,0.0)),List(Rule(Set(140855),List(ItemScore(368788,0.0,1.3516984090509725E-5,0.1111111111111111,38.59207094418362)))))
像这样尝试列表连接++
case class ItemScore(i: Int)
case class Rule(s: String, itemScores: List[ItemScore])
case class PredictedResult(itemScores: List[ItemScore], rules: List[Rule])
val pr1 = PredictedResult(List(ItemScore(1)), Nil)
val pr2 = PredictedResult(Nil, List(Rule("rule1", List(ItemScore(2)))))
PredictedResult(pr1.itemScores ++ pr2.itemScores, pr1.rules ++ pr2.rules)
输出
res0: PredictedResult = PredictedResult(List(ItemScore(1)),List(Rule(rule1,List(ItemScore(2)))))