使用树输出在Spark中使用渐变增强树来预测类的概率

Question

众所周知，Spark中的GBT为您提供了截至目前的预测标签。

我正在考虑计算一个类的预测概率（比如所有实例都属于某个叶子）

构建GBT的代码

import org.apache.spark.SparkContext
import org.apache.spark.mllib.regression.LabeledPoint
import org.apache.spark.mllib.linalg.Vectors
import org.apache.spark.mllib.tree.GradientBoostedTrees
import org.apache.spark.mllib.tree.configuration.BoostingStrategy
import org.apache.spark.mllib.tree.model.GradientBoostedTreesModel
import org.apache.spark.mllib.util.MLUtils

//Importing the data
val data = sc.textFile("data/mllib/credit_approval_2_attr.csv") //using the credit approval data set from UCI machine learning repository

//Parsing the data
val parsedData = data.map { line =>
    val parts = line.split(',').map(_.toDouble)
    LabeledPoint(parts(0), Vectors.dense(parts.tail))
}

//Splitting the data
val splits = parsedData.randomSplit(Array(0.7, 0.3), seed = 11L)
val training = splits(0).cache() 
val test = splits(1)

// Train a GradientBoostedTrees model.
// The defaultParams for Classification use LogLoss by default.
val boostingStrategy = BoostingStrategy.defaultParams("Classification")
boostingStrategy.numIterations = 2 // We can use more iterations in practice.
boostingStrategy.treeStrategy.numClasses = 2
boostingStrategy.treeStrategy.maxDepth = 2
boostingStrategy.treeStrategy.maxBins = 32
boostingStrategy.treeStrategy.subsamplingRate = 0.5
boostingStrategy.treeStrategy.maxMemoryInMB =1024
boostingStrategy.learningRate = 0.1

// Empty categoricalFeaturesInfo indicates all features are continuous.
boostingStrategy.treeStrategy.categoricalFeaturesInfo = Map[Int, Int]()

val model = GradientBoostedTrees.train(training, boostingStrategy)  

model.toDebugString

为了简单起见，这给了我2棵深度为2的树：

 Tree 0:
    If (feature 3 <= 2.0)
     If (feature 2 <= 1.25)
      Predict: -0.5752212389380531
     Else (feature 2 > 1.25)
      Predict: 0.07462686567164178
    Else (feature 3 > 2.0)
     If (feature 0 <= 30.17)
      Predict: 0.7272727272727273
     Else (feature 0 > 30.17)
      Predict: 1.0
  Tree 1:
    If (feature 5 <= 67.0)
     If (feature 4 <= 100.0)
      Predict: 0.5739387416147804
     Else (feature 4 > 100.0)
      Predict: -0.550117566730937
    Else (feature 5 > 67.0)
     If (feature 2 <= 0.0)
      Predict: 3.0383669122382835
     Else (feature 2 > 0.0)
      Predict: 0.4332824083446489

我的问题是：我可以使用上面的树来计算预测概率，如：

关于用于预测的特征集中的每个实例

exp（来自树0的叶子得分+来自树1的叶子得分）/（1 + exp（来自树0的叶子得分+来自树1的叶子得分））

这给了我一种概率。但不确定这是否是正确的方法。此外，如果有任何文件解释如何计算叶子得分（预测）。如果有人可以分享，我将非常感激。

任何建议都是精湛的。

Answer 1

这是我使用Spark内部依赖项的方法。您需要稍后为矩阵运算导入线性代数库，即将树预测乘以学习率。

import org.apache.spark.mllib.linalg.{Vectors, Matrices}
import org.apache.spark.mllib.linalg.distributed.{RowMatrix}

假设您使用GBT构建模型：

val model = GradientBoostedTrees.train(trainingData, boostingStrategy)

使用模型对象计算概率：

// Get the log odds predictions from each tree
val treePredictions = testData.map { point => model.trees.map(_.predict(point.features)) }

// Transform the arrays into matrices for multiplication
val treePredictionsVector = treePredictions.map(array => Vectors.dense(array))
val treePredictionsMatrix = new RowMatrix(treePredictionsVector)
val learningRate = model.treeWeights
val learningRateMatrix = Matrices.dense(learningRate.size, 1, learningRate)
val weightedTreePredictions = treePredictionsMatrix.multiply(learningRateMatrix)

// Calculate probability by ensembling the log odds
val classProb = weightedTreePredictions.rows.flatMap(_.toArray).map(x => 1 / (1 + Math.exp(-1 * x)))
classProb.collect

// You may tweak your decision boundary for different class labels
val classLabel = classProb.map(x => if (x > 0.5) 1.0 else 0.0)
classLabel.collect

以下是您可以复制的代码段＆amp;直接粘贴到spark-shell中：

import org.apache.spark.mllib.regression.LabeledPoint
import org.apache.spark.mllib.linalg.{Vectors, Matrices}
import org.apache.spark.mllib.linalg.distributed.{RowMatrix}
import org.apache.spark.mllib.tree.GradientBoostedTrees
import org.apache.spark.mllib.tree.configuration.BoostingStrategy
import org.apache.spark.mllib.tree.model.GradientBoostedTreesModel

// Load and parse the data file.
val csvData = sc.textFile("data/mllib/sample_tree_data.csv")
val data = csvData.map { line =>
  val parts = line.split(',').map(_.toDouble)
  LabeledPoint(parts(0), Vectors.dense(parts.tail))
}
// Split the data into training and test sets (30% held out for testing)
val splits = data.randomSplit(Array(0.7, 0.3))
val (trainingData, testData) = (splits(0), splits(1))

// Train a GBT model.
val boostingStrategy = BoostingStrategy.defaultParams("Classification")
boostingStrategy.numIterations = 50
boostingStrategy.treeStrategy.numClasses = 2
boostingStrategy.treeStrategy.maxDepth = 6
boostingStrategy.treeStrategy.categoricalFeaturesInfo = Map[Int, Int]()

val model = GradientBoostedTrees.train(trainingData, boostingStrategy)

// Get class label from raw predict function
val predictedLabels = model.predict(testData.map(_.features))
predictedLabels.collect

// Get class probability
val treePredictions = testData.map { point => model.trees.map(_.predict(point.features)) }
val treePredictionsVector = treePredictions.map(array => Vectors.dense(array))
val treePredictionsMatrix = new RowMatrix(treePredictionsVector)
val learningRate = model.treeWeights
val learningRateMatrix = Matrices.dense(learningRate.size, 1, learningRate)
val weightedTreePredictions = treePredictionsMatrix.multiply(learningRateMatrix)
val classProb = weightedTreePredictions.rows.flatMap(_.toArray).map(x => 1 / (1 + Math.exp(-1 * x)))
val classLabel = classProb.map(x => if (x > 0.5) 1.0 else 0.0)
classLabel.collect

Answer 2

def score(features: Vector,gbdt: GradientBoostedTreesModel): Double = {
    val treePredictions = gbdt.trees.map(_.predict(features))
    blas.ddot(gbdt.numTrees, treePredictions, 1, gbdt.treeWeights, 1)
}
def sigmoid(v : Double) : Double = {
    1/(1+Math.exp(-v))
}
// model is output of GradientBoostedTrees.train(...,...)
// testData is libSVM format
val labelAndPreds = testData.map { point =>
        var prediction = score(point.features,model)
        prediction = sigmoid(prediction)
        (point.label, Vectors.dense(1.0-prediction, prediction))
}

Answer 3

实际上，我能够使用树和问题中给出的树的公式来预测概率。我实际上检查了GBT预测的标签输出。当我使用阈值为0.5时，它完全匹配。

所以我们做了同样的改变。

关于用于预测的特征集中的每个实例：

exp（来自树0的叶子得分+（learning_rate）*来自树1的叶子得分）/（1 + exp（来自树0的叶子得分+（learning_rate）*来自树1的叶子得分））

这基本上给了我预测的概率。

我在深度为3的3棵树上测试过相同的效果。并且还有不同的数据集。

很高兴知道其他人是否已经尝试过这个。 如果没有，他们可以尝试这个并发表评论。

Answer 4

实际上，上面的ans是错误的，sigmoid函数在这种情况下是假的，因为spark将标签转换为{-1,1}。你应该使用这样的代码：

def score(features: Vector,gbdt: GradientBoostedTreesModel): Double = {
    val treePredictions = gbdt.trees.map(_.predict(features))
    blas.ddot(gbdt.numTrees, treePredictions, 1, gbdt.treeWeights, 1)
}
val labelAndPreds = testData.map { point =>
        var prediction = score(point.features,model)
        prediction = 1.0 / (1.0 + math.exp(-2.0 * prediction))
        (point.label, Vectors.dense(1.0-prediction, prediction))
}

更多细节可以在＆＃34;贪婪函数逼近的第9页中看到？ Gradient Boosting Machine＆＃34;。还有一个关于spark的拉取请求：https://github.com/apache/spark/pull/16441

Answer 5

实际上，@ hbghhy看到的是错误的，@ Run2是正确的，Spark使用二项式对数对数似然率是Loss的两倍，而Friedman使用二项式对数对数似然率是Loss的“贪婪函数近似”的第9页，< / strong>。

/**
 * :: DeveloperApi ::
 * Class for log loss calculation (for classification).
 * This uses twice the binomial negative log likelihood, called "deviance" in Friedman (1999).
 *
 * The log loss is defined as:
 *   2 log(1 + exp(-2 y F(x)))
 * where y is a label in {-1, 1} and F(x) is the model prediction for features x.
 */
@Since("1.2.0")
@DeveloperApi
object LogLoss extends ClassificationLoss {

  /**
   * Method to calculate the loss gradients for the gradient boosting calculation for binary
   * classification
   * The gradient with respect to F(x) is: - 4 y / (1 + exp(2 y F(x)))
   * @param prediction Predicted label.
   * @param label True label.
   * @return Loss gradient
   */
  @Since("1.2.0")
  override def gradient(prediction: Double, label: Double): Double = {
    - 4.0 * label / (1.0 + math.exp(2.0 * label * prediction))
  }

  override private[spark] def computeError(prediction: Double, label: Double): Double = {
    val margin = 2.0 * label * prediction
    // The following is equivalent to 2.0 * log(1 + exp(-margin)) but more numerically stable.
    2.0 * MLUtils.log1pExp(-margin)
  }
}