我的 RDD 带有元组值(String,SparseVector),我想使用 RDD 创建 DataFrame 。获取(标签:字符串,要素:向量) DataFrame ,这是大多数ml算法库所需的模式。 我知道可以这样做,因为HashingTF ml Library在给出 DataFrame 的功能列时会输出一个向量。
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,我能以某种方式将(String,SparseVector)的 RDD 转换为(String,vector)的 DataFrame 。
我尝试了通常的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)
]))
答案 0 :(得分:19)
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)
为了完整性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)
答案 1 :(得分:4)
虽然@ zero323 answer https://stackoverflow.com/a/32745924/1333621有意义,我希望它对我有用 - 数据帧的基础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 - 如果某人有更短/更好的方式我想知道
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|
+------------------+-----+
答案 2 :(得分:1)
这是scala for spark 2.1的一个例子
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
}
编译器在功能rdd
上无法识别toDF()