How do I convert an RDD with a SparseVector Column to a DataFrame with a column as Vector

Question

I have an RDD with a tuple of values (String, SparseVector) and I want to create a DataFrame using the RDD. To get a (labe

Answer 1

You have to use VectorUDT here:

# In Spark 1.x
# from pyspark.mllib.linalg import SparseVector, VectorUDT
from pyspark.ml.linalg import SparseVector, VectorUDT

temp_rdd = sc.parallelize([
    (0.0, SparseVector(4, {1: 1.0, 3: 5.5})),
    (1.0, SparseVector(4, {0: -1.0, 2: 0.5}))])

schema = StructType([
    StructField("label", DoubleType(), True),
    StructField("features", VectorUDT(), True)
])

temp_rdd.toDF(schema).printSchema()

## root
##  |-- label: double (nullable = true)
##  |-- features: vector (nullable = true)

Just for completeness Scala equivalent:

import org.apache.spark.sql.Row
import org.apache.spark.rdd.RDD
import org.apache.spark.sql.types.{DoubleType, StructType}
// In Spark 1x.
// import org.apache.spark.mllib.linalg.{Vectors, VectorUDT}
import org.apache.spark.ml.linalg.Vectors
import org.apache.spark.ml.linalg.SQLDataTypes.VectorType

val schema = new StructType()
  .add("label", DoubleType)
   // In Spark 1.x
   //.add("features", new VectorUDT())
  .add("features",VectorType)

val temp_rdd: RDD[Row]  = sc.parallelize(Seq(
  Row(0.0, Vectors.sparse(4, Seq((1, 1.0), (3, 5.5)))),
  Row(1.0, Vectors.sparse(4, Seq((0, -1.0), (2, 0.5))))
))

spark.createDataFrame(temp_rdd, schema).printSchema

// root
// |-- label: double (nullable = true)
// |-- features: vector (nullable = true)