如何将具有 SparseVector 列的 RDD 转换为具有列作为 Vector 的 DataFrame

Question

我有一个 RDD 和一个值元组（String，SparseVector），我想创建一个 DataFrame 使用 RDD。获取 (label:string, features:vector) DataFrame 这是大多数 ml 算法库所需的架构。我知道这是可以做到的，因为 HashingTF ml 库在给定 DataFrame 的特征列时输出一个向量。

temp_df = sqlContext.createDataFrame(temp_rdd, StructType([
        StructField("label", DoubleType(), False),
        StructField("tokens", ArrayType(StringType()), False)
    ]))

#assumming there is an RDD (double,array(strings))

hashingTF = HashingTF(numFeatures=COMBINATIONS, inputCol="tokens", outputCol="features")

ndf = hashingTF.transform(temp_df)
ndf.printSchema()

#outputs 
#root
#|-- label: double (nullable = false)
#|-- tokens: array (nullable = false)
#|    |-- element: string (containsNull = true)
#|-- features: vector (nullable = true)

所以我的问题是，我能否以某种方式将 (String, SparseVector) 的 RDD 转换为 (String) 的 DataFrame , 矢量). 我尝试使用通常的 sqlContext.createDataFrame，但没有 DataType 符合我的需求。

df = sqlContext.createDataFrame(rdd,StructType([
        StructField("label" , StringType(),True),
        StructField("features" , ?Type(),True)
    ]))

Answer 1

你必须在这里使用 VectorUDT:

# 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)

为了完整起见，Scala 等价物：

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)

Answer 2

虽然 @zero323 的回答是有道理的，但我希望它对我有用——数据框下面的 rdd，sqlContext.createDataFrame(temp_rdd, schema)，仍然包含 SparseVectors类型我必须执行以下操作才能转换为 DenseVector 类型 - 如果有人有 shorter/better 我想知道的方式

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()
df_w_ftr = temp_rdd.toDF(schema)

print 'original convertion method: ',df_w_ftr.take(5)
print('\n')
temp_rdd_dense = temp_rdd.map(lambda x: Row(label=x[0],features=DenseVector(x[1].toArray())))
print type(temp_rdd_dense), type(temp_rdd)
print 'using map and toArray:', temp_rdd_dense.take(5)

temp_rdd_dense.toDF().show()

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

original convertion method:  [Row(label=0.0, features=SparseVector(4, {1: 1.0, 3: 5.5})), Row(label=1.0, features=SparseVector(4, {0: -1.0, 2: 0.5}))]


<class 'pyspark.rdd.PipelinedRDD'> <class 'pyspark.rdd.RDD'>
using map and toArray: [Row(features=DenseVector([0.0, 1.0, 0.0, 5.5]), label=0.0), Row(features=DenseVector([-1.0, 0.0, 0.5, 0.0]), label=1.0)]

+------------------+-----+
|          features|label|
+------------------+-----+
| [0.0,1.0,0.0,5.5]|  0.0|
|[-1.0,0.0,0.5,0.0]|  1.0|
+------------------+-----+

Answer 3

这是 spark 2.1 的 scala 示例

import org.apache.spark.ml.linalg.Vector

def featuresRDD2DataFrame(features: RDD[Vector]): DataFrame = {
    import sparkSession.implicits._
    val rdd: RDD[(Double, Vector)] = features.map(x => (0.0, x))
    val df = rdd.toDF("label","features").select("features")
    df
  }

toDF() 编译器无法识别 rdd

的特性

如何将具有 SparseVector 列的 RDD 转换为具有列作为 Vector 的 DataFrame

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

apache-spark

pyspark

apache-spark-sql

apache-spark-mllib

apache-spark-ml