你可以做手工这样的:
import org.apache.spark.SparkException
import org.apache.spark.ml.linalg.{Vector, Vectors}
import org.apache.spark.sql.SparkSession
import scala.collection.mutable.ArrayBuilder
case class Row(a: Double, b: Option[Double], c: Double, d: Vector, e: Double)
val dataset = spark.createDataFrame(
Seq(new Row(0, None, 3.0, Vectors.dense(4.0, 5.0, 0.5), 7.0),
new Row(1, Some(2.0), 3.0, Vectors.dense(4.0, 5.0, 0.5), 7.0))
).toDF("id", "hour", "mobile", "userFeatures", "clicked")
val sparseVectorRDD = dataset.rdd.map { row =>
val indices = ArrayBuilder.make[Int]
val values = ArrayBuilder.make[Double]
var cur = 0
row.toSeq.foreach {
case v: Double =>
indices += cur
values += v
cur += 1
case vec: Vector =>
vec.foreachActive { case (i, v) =>
indices += cur + i
values += v
}
cur += vec.size
case null =>
cur += 1
case o =>
throw new SparkException(s"$o of type ${o.getClass.getName} is not supported.")
}
Vectors.sparse(cur, indices.result(), values.result())
}
,然后根据需要将其转换回一个数据帧。由于Row对象未经过类型检查,因此必须手动处理并根据需要转换为适当的类型。
听起来不错!非常感谢!!!! –