045 RDD与DataFrame互相转换

 1     import sqlContext.implicits._                //如果不写，下面的转换不成功
 2 
 3     //transform
 4     val path="/spark/logs/input"
 5     val rdd=sc.textFile(path)
 6     val apacheAccessDataFrame=rdd
 7       .filter(line=>ApacheAccessLog.isValidateLogLine(line))
 8       .map(line => {
 9         ApacheAccessLog.parseLogLine(line)
10     }).toDF()                                    //rdd转换为DataFrame

 1 package com.scala.it
 2 import org.apache.spark.rdd.RDD
 3 import org.apache.spark.sql.types._
 4 import org.apache.spark.sql.{Row, SQLContext}
 5 import org.apache.spark.{SparkConf, SparkContext}
 6 object CreateDataFrameDemo {
 7   def main(args: Array[String]): Unit = {
 8     val conf = new SparkConf()
 9       .setMaster("local[*]")
10       .setAppName("hive-join-mysql")
11     // 使用kryo序列化机制
12     conf.registerKryoClasses(Array(classOf[Row], classOf[Tuple3[Int, String, Double]]))
13     val sc = SparkContext.getOrCreate(conf)
14 
15     val sqlContext = new SQLContext(sc)
16 
17     // ===================================
18     // RDD中Row中的各个列的类型必须是一致的(不能有歧义)
19     val rdd: RDD[Row] = sc.parallelize(Array(
20       (1, "Tom", 1234.1),
21       (2, "Lili", 12532.2),
22       (3, "Gerry", 123.0)
23     )).map {
24       case (id, name, salary) => {
25         Row(id, name, salary)
26       }
27     }
28     val schema: StructType = StructType(Array(
29       StructField("id", IntegerType),
30       StructField("name", StringType),
31       StructField("salary", DoubleType)
32     ))
33 
34     val df = sqlContext.createDataFrame(rdd, schema)
35     df.show()
36   }
37 }