将 RDD[org.apache.spark.sql.Row] 转换为 RDD[org.apache.spark.mllib.linalg.Vector] [英] Converting RDD[org.apache.spark.sql.Row] to RDD[org.apache.spark.mllib.linalg.Vector]

问题描述

我对 Spark 和 Scala 比较陌生.

I am relatively new to Spark and Scala.

我从以下数据框开始(由密集的双精度向量组成的单列):

I am starting with the following dataframe (single column made out of a dense Vector of Doubles):

scala> val scaledDataOnly_pruned = scaledDataOnly.select("features")
scaledDataOnly_pruned: org.apache.spark.sql.DataFrame = [features: vector]

scala> scaledDataOnly_pruned.show(5)
+--------------------+
|            features|
+--------------------+
|[-0.0948337274182...|
|[-0.0948337274182...|
|[-0.0948337274182...|
|[-0.0948337274182...|
|[-0.0948337274182...|
+--------------------+

直接转换到 RDD 会产生一个 org.apache.spark.rdd.RDD[org.apache.spark.sql.Row] 的实例:

A straight conversion to RDD yields an instance of org.apache.spark.rdd.RDD[org.apache.spark.sql.Row] :

scala> val scaledDataOnly_rdd = scaledDataOnly_pruned.rdd
scaledDataOnly_rdd: org.apache.spark.rdd.RDD[org.apache.spark.sql.Row] = MapPartitionsRDD[32] at rdd at <console>:66

有谁知道如何将此 DF 转换为 org.apache.spark.rdd.RDD[org.apache.spark.mllib.linalg.Vector] 的实例?到目前为止，我的各种尝试都没有成功.

Does anyone know how to convert this DF to an instance of org.apache.spark.rdd.RDD[org.apache.spark.mllib.linalg.Vector] instead? My various attempts have been unsuccessful so far.

提前感谢您的指点！

推荐答案

刚刚发现:

val scaledDataOnly_rdd = scaledDataOnly_pruned.map{x:Row => x.getAs[Vector](0)}

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