RDD转为Dataset如何指定schema?

与RDD进行互操作

Spark SQL支持两种不同方法将现有RDD转换为Datasets。第一种方法使用反射来推断包含特定类型对象的RDD的schema。这种基于反射的方法会导致更简洁的代码，并且在编写Spark应用程序时已经知道schema的情况下工作良好。

第二种创建Datasets的方法是通过编程接口，允许您构建schema，然后将其应用于现有的RDD。虽然此方法更详细，但它允许你在直到运行时才知道列及其类型的情况下去构件数据集。

使用反射推断模式

Spark SQL的Scala接口支持自动将包含case classes的RDD转换为DataFrame。Case class定义表的schema。使用反射读取case class的参数名称，并将其变为列的名称。Case class也可以嵌套或包含复杂类型，如Seqs或Arrays。此RDD可以隐式转换为DataFrame，然后将其注册为表格。表可以在随后的SQL语句中使用。

// For implicit conversions from RDDs to DataFrames

import spark.implicits._

// Create an RDD of Person objects from a text file, convert it to a Dataframe

val peopleDF = spark.sparkContext

.textFile("examples/src/main/resources/people.txt")

.map(_.split(","))

.map(attributes => Person(attributes(0), attributes(1).trim.toInt))

.toDF()

// Register the DataFrame as a temporary view

peopleDF.createOrReplaceTempView("people")

// SQL statements can be run by using the sql methods provided by Spark

val teenagersDF = spark.sql("SELECT name, age FROM people WHERE age BETWEEN 13 AND 19")

// The columns of a row in the result can be accessed by field index

teenagersDF.map(teenager => "Name: " + teenager(0)).show()

// +------------+

// | value|

// +------------+

// |Name: Justin|

// +------------+

// or by field name

teenagersDF.map(teenager => "Name: " + teenager.getAs[String]("name")).show()

// +------------+

// | value|

// +------------+

// |Name: Justin|

// +------------+

// No pre-defined encoders for Dataset[Map[K,V]], define explicitly

implicit val mapEncoder = org.apache.spark.sql.Encoders.kryo[Map[String, Any]]

// Primitive types and case classes can be also defined as

// implicit val stringIntMapEncoder: Encoder[Map[String, Any]] = ExpressionEncoder()

// row.getValuesMap[T] retrieves multiple columns at once into a Map[String, T]

teenagersDF.map(teenager => teenager.getValuesMap[Any](List("name", "age"))).collect()

// Array(Map("name" -> "Justin", "age" -> 19))

以编程方式指定模式

当case class不能提前定义时（例如，记录的结构用字符串编码，或者文本数据集将被解析并且字段对不同的用户值会不同），DataFrame可以以编程方式通过三个步骤创建 。

1， Row从原始RDD 创建元素类型为Row的RDD;

2，使用StructType创建一组schema，然后让其匹配步骤1中Rows的类型结构。

3，使用SparkSession 提供的方法createDataFrame，将schema应用于Rows 类型的RDD。

import org.apache.spark.sql.types._

// Create an RDD

val peopleRDD = spark.sparkContext.textFile("examples/src/main/resources/people.txt")

// The schema is encoded in a string

val schemaString = "name age"

// Generate the schema based on the string of schema

val fields = schemaString.split(" ")

.map(fieldName => StructField(fieldName, StringType, nullable = true))

val schema = StructType(fields)

// Convert records of the RDD (people) to Rows

val rowRDD = peopleRDD

.map(_.split(","))

.map(attributes => Row(attributes(0), attributes(1).trim))

// Apply the schema to the RDD

val peopleDF = spark.createDataFrame(rowRDD, schema)

// Creates a temporary view using the DataFrame

peopleDF.createOrReplaceTempView("people")

// SQL can be run over a temporary view created using DataFrames

val results = spark.sql("SELECT name FROM people")

// The results of SQL queries are DataFrames and support all the normal RDD operations

// The columns of a row in the result can be accessed by field index or by field name

results.map(attributes => "Name: " + attributes(0)).show()

// +-------------+

// | value|

// +-------------+

// |Name: Michael|

// | Name: Andy|

// | Name: Justin|

// +-------------+