apache-spark - 以编程方式为 Apache Spark 中的数据框生成模式和数据

Question

我想动态生成一个包含报告标题记录的数据框，因此根据以下字符串的值创建一个数据框:

val headerDescs : String = "Name,Age,Location"

val headerSchema = StructType(headerDescs.split(",").map(fieldName => StructField(fieldName, StringType, true)))

但是现在我想对数据做同样的事情(实际上是相同的数据，即元数据)。

我创建一个 RDD:

val headerRDD = sc.parallelize(headerDescs.split(","))

然后我打算使用 createDataFrame 来创建它:

val headerDf = sqlContext.createDataFrame(headerRDD, headerSchema)

但是失败了，因为 createDataframe 需要一个 RDD[Row]，但是我的 RDD 是一个字符串数组 - 我找不到转换我的 RDD 的方法到 Row RDD，然后动态映射字段。我见过的示例假设您事先知道列数，但我希望最终能够在不更改代码的情况下更改列 - 例如，将列放在文件中。

基于第一个答案的代码摘录:

val headerDescs : String = "Name,Age,Location"

// create the schema from a string, splitting by delimiter
val headerSchema = StructType(headerDescs.split(",").map(fieldName => StructField(fieldName, StringType, true)))

// create a row from a string, splitting by delimiter
val headerRDDRows = sc.parallelize(headerDescs.split(",")).map( a => Row(a))

val headerDf = sqlContext.createDataFrame(headerRDDRows, headerSchema)
headerDf.show()

执行此结果:

+--------+---+--------+

|    Name|Age|Location|

+--------+---+--------+

|    Name|

|     Age|

|Location|

+--------+---+-------

Answer 1

要将 RDD[Array[String]] 转换为 RDD[Row]，您需要执行以下步骤:

导入 org.apache.spark.sql.Row

val headerRDD = sc.parallelize(Seq(headerDescs.split(","))).map(x=>Row(x(0),x(1),x(2)))

scala> val headerSchema = StructType(headerDescs.split(",").map(fieldName => StructField(fieldName, StringType, true)))
headerSchema: org.apache.spark.sql.types.StructType = StructType(StructField(Name,StringType,true), StructField(Age,StringType,true), StructField(Location,StringType,true))

scala> val headerRDD = sc.parallelize(Seq(headerDescs.split(","))).map(x=>Row(x(0),x(1),x(2)))
headerRDD: org.apache.spark.rdd.RDD[org.apache.spark.sql.Row] = MapPartitionsRDD[6] at map at <console>:34

scala> val headerDf = sqlContext.createDataFrame(headerRDD, headerSchema)
headerDf: org.apache.spark.sql.DataFrame = [Name: string, Age: string, Location: string]


scala> headerDf.printSchema
root
 |-- Name: string (nullable = true)
 |-- Age: string (nullable = true)
 |-- Location: string (nullable = true)



scala> headerDf.show
+----+---+--------+
|Name|Age|Location|
+----+---+--------+
|Name|Age|Location|
+----+---+--------+

这会给你一个 RDD[Row]

For reading through file

val vRDD = sc.textFile("..**filepath**.").map(_.split(",")).map(a => Row.fromSeq(a))
 
val headerDf = sqlContext.createDataFrame(vRDD , headerSchema)

Using Spark-CSV package :

 val df = sqlContext.read
    .format("com.databricks.spark.csv")
    .option("header", "true") // Use first line of all files as header
    .schema(headerSchema) // defining based on the custom schema
    .load("cars.csv")

或

val df = sqlContext.read
    .format("com.databricks.spark.csv")
    .option("header", "true") // Use first line of all files as header
    .option("inferSchema", "true") // Automatically infer data types
    .load("cars.csv")

您还可以在其 documentation 中探索各种选项。 .