aggregate
package spark.examples.rddapi import org.apache.spark.{SparkConf, SparkContext} //测试RDD的aggregate方法 object AggregateTest { def main(args: Array[String]) { val conf = new SparkConf().setMaster("local").setAppName("AggregateTest_00") val sc = new SparkContext(conf); val z1 = sc.parallelize(List(1, 3, 5, 7, 7, 5, 3, 3, 79), 2) /** * Aggregate the elements of each partition, and then the results for all the partitions, using * given combine functions and a neutral "zero value". This function can return a different result * type, U, than the type of this RDD, T. Thus, we need one operation for merging a T into an U * and one operation for merging two U's, as in scala.TraversableOnce. Both of these functions are * allowed to modify and return their first argument instead of creating a new U to avoid memory * allocation. */ // def aggregate[U: ClassTag](zeroValue: U)(seqOp: (U, T) => U, combOp: (U, U) => U): U //T是RDD中的元素类型,U是aggregate方法自定义的泛型参数,aggregate返回U(而不一定是T) //两个分区取最大值,然后相加 //math.max(_, _)表示针对每个partition实施的操作, _ + _表示combiner val r1 = z1.aggregate(0)(math.max(_, _), _ + _) println(r1) //86 //RDD元素类型字符串,aggregate的返回类型同样为String val z2 = sc.parallelize(List("a", "b", "c", "d", "e", "f"), 2) val r2 = z2.aggregate("xx")(_ + _, _ + _) println(r2) //连接操作,结果xxxxabcxxdef,每个分区计算时,加上xx,最后两个分区计算时,继续把xx加上 //_ + _的道理也是(x,y) => x + y //(x,y)=>math.max是做两两比较吗? val z3 = sc.parallelize(List("12", "23", "345", "4567"), 2) val r3 = z3.aggregate("")((x, y) => math.max(x.length, y.length).toString, (x, y) => x + y) println(r3) ///结果24,表示两个分区的字符串长度最长的长度转成String后,做拼接 //结果为什么是11? val r4 = sc.parallelize(List("12", "23", "345", "4567"), 2).aggregate("")((x, y) => math.min(x.length, y.length).toString, (x, y) => x + y) println(r4) } }
cartesian
package spark.examples.rddapi import org.apache.spark.rdd.{CartesianRDD, RDD} import org.apache.spark.{SparkContext, SparkConf} object CartesianTest_01 { def main(args: Array[String]) { val conf = new SparkConf().setMaster("local").setAppName("AggregateTest_00") val sc = new SparkContext(conf); val z1 = sc.parallelize(List(2, 3, 4, 5, 6), 2) val z2 = sc.parallelize(List("A", "B", "C", "D", "E", "F", "G", "H", "I", "J"), 3) /** * Return the Cartesian product of this RDD and another one, that is, the RDD of all pairs of * elements (a, b) where a is in `this` and b is in `other`. */ //def cartesian[U: ClassTag](other: RDD[U]): RDD[(T, U)] = new CartesianRDD(sc, this, other) //z1 和 z2集合的元素类型可以不同,并且cartesian是个转换算子, //调用z.collect触发作业 val z = z1.cartesian(z2) println("Number of partitions: " + z.partitions.length) //6 var count = 0 z.collect().foreach(x => {println(x._1 + "," + x._2); count = count + 1}) // println("count =" + count) //50
checkpoint
注意点:
Checkpointed RDDs are stored as a binary file within the checkpoint directory which can be specied using the Spark context. (Warning: Spark applies lazy evaluation. Checkpointing will not occur until an action is invoked.) Important note: the directory "my directory name" should exist in all slaves. As an alternative you could use an HDFS directory URL as well.
package spark.examples.rddapi import org.apache.spark.{SparkContext, SparkConf} object CheckpointTest { def main(args: Array[String]) { val conf = new SparkConf().setMaster("local").setAppName("AggregateTest_00") val sc = new SparkContext(conf); val z = sc.parallelize(List(3, 6, 7, 9, 11)) sc.setCheckpointDir("file:///d:/checkpoint") /** * Mark this RDD for checkpointing. It will be saved to a file inside the checkpoint * directory set with SparkContext.setCheckpointDir() and all references to its parent * RDDs will be removed. This function must be called before any job has been * executed on this RDD. It is strongly recommended that this RDD is persisted in * memory, otherwise saving it on a file will require recomputation. */ z.checkpoint() println("length: " + z.collect().length) //rdd存入目录 println("count: " + z.count()) //5 } }
d:\checkpoint>tree /f 文件夹 PATH 列表 卷序列号为 EA23-0890 D:. └─9b0ca0d9-f7fb-46bb-84dc-097d95b9e7b8 └─rdd-0 .part-00000.crc part-00000
1. 运行过程中发现,checkpoint目录会自动创建,无需预创建
2.程序运行结束后,checkpoint目录并没有删除,上面这些属于checkpoint目录下的目录和文件也没有删除,再次运行会产生新的目录
Repartition
package spark.examples.rddapi import org.apache.spark.{SparkContext, SparkConf} object RepartitionTest_04 { def main(args: Array[String]) { val conf = new SparkConf().setMaster("local").setAppName("RepartitionTest_04") val sc = new SparkContext(conf); val z1 = sc.parallelize(List(3, 9, 18, 22, 11, 9, 8), 3) //z1.coalesce(5, true)的效果一样,开启shuffle /** * Return a new RDD that has exactly numPartitions partitions. * * Can increase or decrease the level of parallelism in this RDD. Internally, this uses * a shuffle to redistribute data. * * If you are decreasing the number of partitions in this RDD, consider using `coalesce`, * which can avoid performing a shuffle. */ val r1 = z1.repartition(5) r1.collect().foreach(println) } }
coalesce
package spark.examples.rddapi import org.apache.spark.{SparkContext, SparkConf} //coalesce:合并 object CoalesceTest_03 { def main(args: Array[String]) { val conf = new SparkConf().setMaster("local").setAppName("CoalesceTest_03") val sc = new SparkContext(conf); val z = sc.parallelize(List(3, 9, 18, 22, 11, 9, 8), 3) /** * Return a new RDD that is reduced into `numPartitions` partitions. * * This results in a narrow dependency, e.g. if you go from 1000 partitions * to 100 partitions, there will not be a shuffle, instead each of the 100 * new partitions will claim 10 of the current partitions. * * However, if you're doing a drastic coalesce, e.g. to numPartitions = 1, * this may result in your computation taking place on fewer nodes than * you like (e.g. one node in the case of numPartitions = 1). To avoid this, * you can pass shuffle = true. This will add a shuffle step, but means the * current upstream partitions will be executed in parallel (per whatever * the current partitioning is). * * Note: With shuffle = true, you can actually coalesce to a larger number * of partitions. This is useful if you have a small number of partitions, * say 100, potentially with a few partitions being abnormally large. Calling * coalesce(1000, shuffle = true) will result in 1000 partitions with the * data distributed using a hash partitioner. */ //shuffle默认为false //将分区数由3变成2,大变小使用narrow dependency val zz = z.coalesce(2, false) println("Partitions length: " + zz.partitions.length) //2 println(zz.collect()) //结果是[I@100498c? zz.collect().foreach(println) //将分区数由3变成6,少变多必须使用shuffle=true //在单机上没有发现有问题 //在cluster环境下,为了保证新的分区分布到不同的节点,应该使用shuffle为true //也就是说,少变多也可以使用shuffle为false,但是达不到分区数据进行重新分布的目的 val z2 = z.coalesce(6, false) z2.collect().foreach(println) //分区扩大,同时设置shuffle为true val z3 = z.coalesce(6, true) z3.collect().foreach(println) } }