RDD到LabeledPoint的转换

Question

如果我有一个大约有500列和2亿行的RDD,并且RDD.columns.indexOf("target", 0)显示Int = 77了我的目标因变量是在第77列.但我对如何选择所需(部分)列作为特征没有足够的知识(比如说)我想要23到59,111到357,399到489的列.我想知道我是否可以申请:

val data = rdd.map(col => new LabeledPoint(
    col(77).toDouble, Vectors.dense(??.map(x => x.toDouble).toArray))

任何建议或指导将不胜感激.

也许我搞砸了RDD与DataFrame,我可以将RDD转换为DataFrame,.toDF()或者使用DataFrame比RDD更容易实现目标.

Answer 1

我假设您的数据看起来或多或少像这样:

import scala.util.Random.{setSeed, nextDouble}
setSeed(1)

case class Record(
    foo: Double, target: Double, x1: Double, x2: Double, x3: Double)

val rows = sc.parallelize(
    (1 to 10).map(_ => Record(
        nextDouble, nextDouble, nextDouble, nextDouble, nextDouble
   ))
)
val df = sqlContext.createDataFrame(rows)
df.registerTempTable("df")

sqlContext.sql("""
  SELECT ROUND(foo, 2) foo,
         ROUND(target, 2) target,
         ROUND(x1, 2) x1,
         ROUND(x2, 2) x2,
         ROUND(x2, 2) x3 
  FROM df""").show

所以我们有如下数据:

+----+------+----+----+----+
| foo|target|  x1|  x2|  x3|
+----+------+----+----+----+
|0.73|  0.41|0.21|0.33|0.33|
|0.01|  0.96|0.94|0.95|0.95|
| 0.4|  0.35|0.29|0.51|0.51|
|0.77|  0.66|0.16|0.38|0.38|
|0.69|  0.81|0.01|0.52|0.52|
|0.14|  0.48|0.54|0.58|0.58|
|0.62|  0.18|0.01|0.16|0.16|
|0.54|  0.97|0.25|0.39|0.39|
|0.43|  0.23|0.89|0.04|0.04|
|0.66|  0.12|0.65|0.98|0.98|
+----+------+----+----+----+

我们要忽略foo和x2和提取LabeledPoint(target, Array(x1, x3)):

// Map feature names to indices
val featInd = List("x1", "x3").map(df.columns.indexOf(_))

// Or if you want to exclude columns
val ignored = List("foo", "target", "x2")
val featInd = df.columns.diff(ignored).map(df.columns.indexOf(_))

// Get index of target
val targetInd = df.columns.indexOf("target") 

df.rdd.map(r => LabeledPoint(
   r.getDouble(targetInd), // Get target value
   // Map feature indices to values
   Vectors.dense(featInd.map(r.getDouble(_)).toArray) 
))