Apache spark Spark:如何获得贝努利朴素贝叶斯的概率和AUC?
我正在运行一个Apache spark Spark:如何获得贝努利朴素贝叶斯的概率和AUC?,apache-spark,pyspark,apache-spark-mllib,naivebayes,apache-spark-ml,Apache Spark,Pyspark,Apache Spark Mllib,Naivebayes,Apache Spark Ml,我正在运行一个Bernoulli Naive Bayes使用代码: val splits = MyData.randomSplit(Array(0.75, 0.25), seed = 2L) val training = splits(0).cache() val test = splits(1) val model = NaiveBayes.train(training, lambda = 3.0, modelType = "bernoulli") 我的问题是如何获得0类(或1类)的成员概率
Bernoulli Naive Bayes
使用代码:
val splits = MyData.randomSplit(Array(0.75, 0.25), seed = 2L)
val training = splits(0).cache()
val test = splits(1)
val model = NaiveBayes.train(training, lambda = 3.0, modelType = "bernoulli")
我的问题是如何获得0类(或1类)的成员概率并计算AUC。我想得到与我使用此代码的地方的LogisticRegressionWithSGD
或SVMWithSGD
类似的结果:
val numIterations = 100
val model = SVMWithSGD.train(training, numIterations)
model.clearThreshold()
// Compute raw scores on the test set.
val labelAndPreds = test.map { point =>
val prediction = model.predict(point.features)
(prediction, point.label)
}
// Get evaluation metrics.
val metrics = new BinaryClassificationMetrics(labelAndPreds)
val auROC = metrics.areaUnderROC()
不幸的是,该代码不适用于
NaiveBayes
关于Bernouilli NaiveBayes的概率,下面是一个示例:
//构建虚拟数据
val data=sc.parallelize(列表(“0,100”、“1,01”、“1,01”、“0,01”、“1,110”))
//将虚拟数据转换为标签点
val parsedData=data.map{line=>
val parts=line.split(',')
标签点(部分(0).toDouble,向量.dense(部分(1).split(“”).map(u.toDouble)))
}
//为培训准备数据
val splits=parsedData.randomspilt(数组(0.75,0.25),种子=2L)
val training=splits(0).cache()
val测试=拆分(1)
val model=NaiveBayes.train(training,lambda=3.0,modelType=“bernoulli”)
//标签
val标签=model.labels
//所有特征向量的概率
val features=parsedData.map(lp=>lp.features)
模型。可预测概率(特征)。每个println取(10)
//对于一个特定的向量,我取parsedData中的第一个向量
val testVector=parsedData.first.features
println(s“对于向量${testVector}=>概率:${model.predictProbability(testVector)}”)
至于AUC:
// Compute raw scores on the test set.
val labelAndPreds = test.map { point =>
val prediction = model.predict(point.features)
(prediction, point.label)
}
// Get evaluation metrics.
val metrics = new BinaryClassificationMetrics(labelAndPreds)
val auROC = metrics.areaUnderROC()
关于聊天室的询问:
val results=parsedData.map{lp=>
val概率:向量=模型预测概率(lp特征)
(对于(i值概率:向量=模型预测概率(lp.特征)
val bestClass=probs.argmax
(标签(最佳等级)、probs(最佳等级))
}
结果:每次打印10次
// (0.0,0.59728640251696)
// (1.0,0.745312681961104)
// (1.0,0.5291306032812298)
// (0.0,0.6496075621805428)
// (1.0,0.5841414717626924)
注意:适用于火花1.5+
编辑:(适用于Pyspark用户)
似乎有些人在使用pyspark和mllib获取概率时遇到了问题。这很正常,spark mllib没有为pyspark提供该函数
因此,您需要使用基于数据帧的API:
from pyspark.sql import Row
from pyspark.ml.linalg import Vectors
from pyspark.ml.classification import NaiveBayes
df = spark.createDataFrame([
Row(label=0.0, features=Vectors.dense([0.0, 0.0])),
Row(label=0.0, features=Vectors.dense([0.0, 1.0])),
Row(label=1.0, features=Vectors.dense([1.0, 0.0]))])
nb = NaiveBayes(smoothing=1.0, modelType="bernoulli")
model = nb.fit(df)
model.transform(df).show(truncate=False)
# +---------+-----+-----------------------------------------+----------------------------------------+----------+
# |features |label|rawPrediction |probability |prediction|
# +---------+-----+-----------------------------------------+----------------------------------------+----------+
# |[0.0,0.0]|0.0 |[-1.4916548767777167,-2.420368128650429] |[0.7168141592920354,0.28318584070796465]|0.0 |
# |[0.0,1.0]|0.0 |[-1.4916548767777167,-3.1135153092103742]|[0.8350515463917526,0.16494845360824742]|0.0 |
# |[1.0,0.0]|1.0 |[-2.5902671654458262,-1.7272209480904837]|[0.29670329670329676,0.7032967032967034]|1.0 |
# +---------+-----+-----------------------------------------+----------------------------------------+----------+
您只需要选择预测列并计算AUC
有关spark ml中朴素贝叶斯的更多信息,请参阅官方文档。Ok这是一个二合一的问题。那么您使用的是哪一版本的spark?您还想要什么的概率?spark 1.5.0。我想要
p(Y=0 | X)
,有了这个,我就可以计算AUC了,对吗?是的,这是一个二进制分类我在使用spark.mllib非常感谢!我对它做了一些修改,现在我可以得到(标签,P(y=0 | x))
:val results=test.map{lp=>val probs:Vector=model.predictabilities(lp.features)val MyList=列表范围(0,(probs.size-1),2)(用于(i