Question

我有一个[vector]类型的列，我有空值，我无法摆脱，这里是一个例子

import org.apache.spark.mllib.linalg.Vectors

val sv1: Vector = Vectors.sparse(58, Array(8, 45), Array(1.0, 1.0))
val df_1 = sc.parallelize(List(("id_1", sv1))).toDF("id", "feature_vector")
val df_2 = sc.parallelize(List(("id_1", 10.0), ("id_2", 10.0))).toDF("id", "numeric_feature")

val df_joined = df_1.join(df_2, Seq("id"), "right")

df_joined.show()

+----+--------------------+---------------+
|  id|      feature_vector|numeric_feature|
+----+--------------------+---------------+
|id_1|(58,[8,45],[1.0,1...|           10.0|
|id_2|                null|           10.0|
+----+--------------------+---------------+

我想做什么：

val map = Map("feature_vector" -> sv1)
val result = df_joined.na.fill(map)

但这会引发错误：

Message: Unsupported value type org.apache.spark.mllib.linalg.SparseVector ((58,[8,45],[1.0,1.0])).

我试过的其他事情：

df_joined.withColumn("feature_vector", when(col("feature_vector").isNull, sv1).otherwise(sv1)).show

来自how to filter out a null value from spark dataframe

我很难找到适用于Spark 1.6的解决方案

Answer 1

合并和加入应该做的伎俩

import org.apache.spark.sql.functions.{coalesce, broadcast}

val fill = Seq(
  Tuple1(Vectors.sparse(58, Array(8, 45), Array(1.0, 1.0)))
).toDF("fill")


df_joined
  .join(broadcast(fill))
  .withColumn("feature_vector", coalesce($"feature_vector", $"fill"))
  .drop("fill")

Answer 2

如果您愿意，可以在此处获取RDD的帮助：

val naFillRDD = df_joined.map{ r => r match{
  case Row(id, feature_vector: Vector, numeric_feature ) => Row(id, feature_vector, numeric_feature )
  case Row(id, _, numeric_feature) => Row(id, sv1, numeric_feature)
}}

然后切换回dataframe：

val naFillDF = sqlContext.createDataFrame(naFillRDD, df_joined.schema)

naFillDF.show(false)
//+----+---------------------+---------------+
//|id  |feature_vector       |numeric_feature|
//+----+---------------------+---------------+
//|id_1|(58,[8,45],[1.0,1.0])|10.0           |
//|id_2|(58,[8,45],[1.0,1.0])|10.0           |
//+----+---------------------+---------------+

如何替换Vector列中的空值？

2 个答案: