Spark：如果 DataFrame 有架构，DataFrame 如何成为 Dataset[Row]

Question

This article claims that a DataFrame in Spark is equivalent to a Dataset[Row] , but this blog post shows that a DataFrame has a schema.

Take the example in the blog post of converting an RDD to a DataFrame : if DataFrame were the same thing as Dataset[Row] , then converting an RDD to a DataFrame should be as simple

val rddToDF = rdd.map(value => Row(value))

But instead it shows that it's this

val rddStringToRowRDD = rdd.map(value => Row(value))
val dfschema = StructType(Array(StructField("value",StringType)))
val rddToDF = sparkSession.createDataFrame(rddStringToRowRDD,dfschema)
val rDDToDataSet = rddToDF.as[String]

Clearly a dataframe is actually a dataset of rows and a schema .

Answer 1

In Spark 2.0, in code there is: type DataFrame = Dataset[Row]

It is Dataset[Row] , just because of definition.

Dataset has also schema, you can print it using printSchema() function. Normally Spark infers schema, so you don't have to write it by yourself - however it's still there ;)

You can also do createTempView(name) and use it in SQL queries, just like DataFrames.

In other words, Dataset = DataFrame from Spark 1.5 + encoder , that converts rows to your classes. After merging types in Spark 2.0, DataFrame becomes just an alias for Dataset[Row] , so without specified encoder.

About conversions: rdd.map() also returns RDD , it never returns DataFrame. You can do:

// Dataset[Row]=DataFrame, without encoder
val rddToDF = sparkSession.createDataFrame(rdd)
// And now it has information, that encoder for String should be used - so it becomes Dataset[String]
val rDDToDataSet = rddToDF.as[String]

// however, it can be shortened to:
val dataset = sparkSession.createDataset(rdd)

Answer 2

Note (in addition to the answer of T Gaweda ) that there is a schema associated to each Row ( Row.schema ). However, this schema is not set until it is integrated in a DataFrame (or Dataset[Row] )

scala> Row(1).schema
res12: org.apache.spark.sql.types.StructType = null

scala> val rdd = sc.parallelize(List(Row(1)))
rdd: org.apache.spark.rdd.RDD[org.apache.spark.sql.Row] = ParallelCollectionRDD[5] at parallelize at <console>:28
scala> spark.createDataFrame(rdd,schema).first
res15: org.apache.spark.sql.Row = [1]
scala> spark.createDataFrame(rdd,schema).first.schema
res16: org.apache.spark.sql.types.StructType = StructType(StructField(a,IntegerType,true))