Scala – 如何将我们在将GMM模型拟合到数据时获得的概率列(向量

发布时间：2020-12-16 18:09:36 所属栏目：安全来源：网络整理

导读：参见英文答案 Spark Scala: How to convert Dataframe[vector] to DataFrame[f1:Double,…,fn: Double)]????????????????????????????????????2个我正在尝试执行以下操作： +-----+-------------------------+----------+---------------------------------

参见英文答案 > Spark Scala: How to convert Dataframe[vector] to DataFrame[f1:Double,…,fn: Double)]????????????????????????????????????2个
我正在尝试执行以下操作：

+-----+-------------------------+----------+-------------------------------------------+
|label|features                 |prediction|probability                                |
+-----+-------------------------+----------+-------------------------------------------+
|0.0  |(3,[],[])                |0         |[0.9999999999999979,2.093996169658831E-15] |
|1.0  |(3,[0,1,2],[0.1,0.1,0.1])|0         |[0.999999999999999,9.891337521299582E-16]  |
|2.0  |(3,[0.2,0.2,0.2])|0         |[0.9999999999999979,2.0939961696578572E-15]|
|3.0  |(3,[9.0,9.0,9.0])|1         |[2.093996169659668E-15,0.9999999999999979] |
|4.0  |(3,[9.1,9.1,9.1])|1         |[9.89133752128275E-16,0.999999999999999]   |
|5.0  |(3,[9.2,9.2,9.2])|1         |[2.0939961696605603E-15,0.9999999999999979]|
+-----+-------------------------+----------+-------------------------------------------+

将上面的数据帧转换为另外两列：prob1& prob2
每列具有在概率列中存在的对应值.

我发现了类似的问题 – 一个在PySpark,另一个在Scala.我不知道如何翻译PySpark代码,我收到了Scala代码的错误.

PySpark代码：

split1_udf = udf(lambda value: value[0].item(),FloatType())
split2_udf = udf(lambda value: value[1].item(),FloatType())

output2 = randomforestoutput.select(split1_udf('probability').alias('c1'),split2_udf('probability').alias('c2'))

或者将这些列附加到原始数据框：

randomforestoutput.withColumn('c1',split1_udf('probability')).withColumn('c2',split2_udf('probability'))

Scala代码：

import org.apache.spark.sql.functions.udf

val getPOne = udf((v: org.apache.spark.mllib.linalg.Vector) => v(1))
model.transform(testDf).select(getPOne($"probability"))

运行Scala代码时出现以下错误：

scala> predictions.select(getPOne(col("probability"))).show(false)
org.apache.spark.sql.AnalysisException: cannot resolve 'UDF(probability)' due to data type mismatch: argument 1 requires vector type,however,'`probability`' is of vector type.;;
'Project [UDF(probability#39) AS UDF(probability)#135]
+- Project [label#0,features#1,prediction#34,UDF(features#1) AS probability#39]
   +- Project [label#0,UDF(features#1) AS prediction#34]
      +- Relation[label#0,features#1] libsvm

我目前正在使用Scala 2.11.11和Spark 2.1.1

解决方法

我从你的问题中理解的是,你试图将概率列分成两列prob1和prob2.如果是这种情况,那么带有withColumn的简单数组功能应该可以解决您的问题.

predictions
  .withColumn("prob1",$"probability"(0))
  .withColumn("prob2",$"probability"(1))
  .drop("probability")

您可以找到可以帮助您将来应用于数据帧的more functions.

编辑

我创建了一个与您的列匹配的临时数据帧

val predictions = Seq(Array(1.0,2.0),Array(2.0939961696605603E-15,0.9999999999999979),Array(Double.NaN,Double.NaN)).toDF("probability")
+--------------------------------------------+
|probability                                 |
+--------------------------------------------+
|[1.0,2.0]                                  |
|[2.0939961696605603E-15,0.9999999999999979]|
|[NaN,NaN]                                  |
+--------------------------------------------+

并且我将上面的结果应用于了

+----------------------+------------------+
|prob1                 |prob2             |
+----------------------+------------------+
|1.0                   |2.0               |
|2.0939961696605603E-15|0.9999999999999979|
|NaN                   |NaN               |
+----------------------+------------------+

架构不匹配编辑

既然由于概率列的Vector模式与上面的arrayType模式解决方案不匹配,上面的解决方案不适用于您的情况.请使用以下解决方案.

您将必须创建udf函数并按预期返回值

val first = udf((v: Vector) => v.toArray(0))
    val second = udf((v: Vector) => v.toArray(1))
    predictions
      .withColumn("prob1",first($"probability"))
      .withColumn("prob2",second($"probability"))
      .drop("probability")

我希望你能得到理想的结果.

（编辑：李大同）

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