Programmatically generate the schema AND the data for a dataframe in Apache Spark

问题

I would like to dynamically generate a dataframe containing a header record for a report, so creating a dataframe from the value of the string below:

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

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

However now I want to do the same for the data (which is in effect the same data i.e. the metadata).

I create an RDD :

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

I then intended to use createDataFrame to create it:

val headerDf = sqlContext.createDataFrame(headerRDD, headerSchema)

however that fails because createDataframe is expecting a RDD[Row], however my RDD is an array of strings - I can't find a way of converting my RDD to a Row RDD and then mapping the fields dynamically. Examples I've seen assume you know the number of columns beforehand, however I want the ability eventually to be able to change the columns without changing the code - having the columns in a file for example.

Code excerpt based on first answer:

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()

Executing this Results in:

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

|    Name|Age|Location|

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

|    Name|

|     Age|

|Location|

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

回答1:

For converting RDD[Array[String]] to RDD[Row] you need to do following steps:

import 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|
+----+---+--------+

This would give you a 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")

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")

There are are various options also which you can explore in its documentation.

来源：https://stackoverflow.com/questions/41745321/programmatically-generate-the-schema-and-the-data-for-a-dataframe-in-apache-spar

标签

apache-spark

dataframe

spark-dataframe

rdd

spark-csv