Ref:http://homepage.cs.latrobe.edu.au/zhe/ZhenHeSparkRDDAPIExamples.html
map是对每个元素操作,
mapPartitions是对其中的每个partition操作
mapPartitionsWithIndex : 把每个partition中的分区号和对应的值拿出来, 看源码
val func = (index: Int, iter: Iterator[(Int)]) => {
iter.toList.map(x => "[partID:" + index + ", val: " + x + "]").iterator
}
def func1(index: Int, iter: Iterator[(Int)]) : Iterator[String] = {
iter.toList.map(x => "[partID:" + index + ", val: " + x + "]").iterator
}
val rdd1 = sc.parallelize(List(1,2,3,4,5,6,7,8,9), 2)
rdd1.mapPartitionsWithIndex(func).collect
rdd1.mapPartitionsWithIndex(func1).collect
#----------------------------------------------------
rdd1.mapPartitionsWithIndex((index, iter) => {
iter.toList.map(x => s"[partID:$index ,val:$x").toIterator
}).collect().foreach(println)aggregate ###是action操作
第一个参数是初始值,
二:是2个函数[每个函数都是2个参数(第一个参数:先对个个分区进行合并,
第二个:对个个分区合并后的结果再进行合并), 输出一个参数]
###0 + (0+1+2+3+4 + 0+5+6+7+8+9)
rdd1.aggregate(0)(_+_, _+_)
rdd1.aggregate(0)(math.max(_, _), _ + _)
###5和1比, 得5再和234比得5 --> 5和6789比,得9 --> 5 + (5+9)
rdd1.aggregate(5)(math.max(_, _), _ + _)
val rdd2 = sc.parallelize(List("a","b","c","d","e","f"),2)
def func2(index: Int, iter: Iterator[(String)]) : Iterator[String] = {
iter.toList.map(x => "[partID:" + index + ", val: " + x + "]").iterator
}
rdd2.aggregate("")(_ + _, _ + _)
rdd2.aggregate("=")(_ + _, _ + _)
val rdd3 = sc.parallelize(List("12","23","345","4567"),2)
rdd3.aggregate("")((x,y) => math.max(x.length, y.length).toString, (x,y) => x + y)
val rdd4 = sc.parallelize(List("12","23","345",""),2)
rdd4.aggregate("")((x,y) => math.min(x.length, y.length).toString, (x,y) => x + y)
val rdd5 = sc.parallelize(List("12","23","","345"),2)
rdd5.aggregate("")((x,y) => math.min(x.length, y.length).toString, (x,y) => x + y)
aggregateByKey
val pairRDD = sc.parallelize(List( ("cat",2), ("cat", 5), ("mouse", 4),("cat", 12), ("dog", 12), ("mouse", 2)), 2)
def func2(index: Int, iter: Iterator[(String, Int)]) : Iterator[String] = {
iter.toList.map(x => "[partID:" + index + ", val: " + x + "]").iterator
}
pairRDD.mapPartitionsWithIndex(func2).collect
pairRDD.aggregateByKey(0)(math.max(_, _), _ + _).collect
pairRDD.aggregateByKey(100)(math.max(_, _), _ + _).collectcombineByKey : 和reduceByKey是相同的效果
###第一个参数x:原封不动取出来,
###第二个参数:是函数, 局部运算,
###第三个:是函数, 对局部运算后的结果再做运算
###每个分区中每个key中value中的第一个值, (hello,1)(hello,1)(good,1)-->(hello(1,1),good(1))-->x就相当于hello的第一个1, good中的1
val rdd1 = sc.textFile("hdfs://master:9000/wordcount/input/").flatMap(_.split(" ")).map((_, 1))
val rdd2 = rdd1.combineByKey(x => x, (a: Int, b: Int) => a + b, (m: Int, n: Int) => m + n)
rdd1.collect
rdd2.collect
combineByKey : 和reduceByKey是相同的效果
###第一个参数x:原封不动取出来, 第二个参数:是函数, 局部运算, 第三个:是函数, 对局部运算后的结果再做运算
###每个分区中每个key中value中的第一个值, (hello,1)(hello,1)(good,1)-->(hello(1,1),good(1))-->x就相当于hello的第一个1, good中的1
val rdd1 = sc.textFile("hdfs://master:9000/wordcount/input/").flatMap(_.split(" ")).map((_, 1))
val rdd2 = rdd1.combineByKey(x => x, (a: Int, b: Int) => a + b, (m: Int, n: Int) => m + n)
rdd1.collect
rdd2.collect
###当input下有3个文件时(有3个block块, 不是有3个文件就有3个block, ), 每个会多加3个10
val rdd3 = rdd1.combineByKey(x => x + 10, (a: Int, b: Int) => a + b, (m: Int, n: Int) => m + n)
rdd3.collect
val rdd4 = sc.parallelize(List("dog","cat","gnu","salmon","rabbit","turkey","wolf","bear","bee"), 3)
val rdd5 = sc.parallelize(List(1,1,2,2,2,1,2,2,2), 3)
val rdd6 = rdd5.zip(rdd4)
val rdd7 = rdd6.combineByKey(List(_), (x: List[String], y: String) => x :+ y, (m: List[String], n: List[String]) => m ++ n)
map是对每个元素操作, mapPartitions是对其中的每个partition操作
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mapPartitionsWithIndex : 把每个partition中的分区号和对应的值拿出来, 看源码
val func = (index: Int, iter: Iterator[(Int)]) => {
iter.toList.map(x => "[partID:" + index + ", val: " + x + "]").iterator
}
val rdd1 = sc.parallelize(List(1,2,3,4,5,6,7,8,9), 2)
rdd1.mapPartitionsWithIndex(func).collect
def func1(index: Int, iter: Iterator[(Int)]) : Iterator[String] = {
iter.toList.map(x => "[partID:" + index + ", val: " + x + "]").iterator
}
val rdd1 = sc.parallelize(List(1,2,3,4,5,6,7,8,9), 2)
rdd1.mapPartitionsWithIndex(func1).collect
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aggregate
###是action操作,
第一个参数是初始值,
二:是2个函数[每个函数都是2个参数(第一个参数:先对个个分区进行合并,
第二个:对个个分区合并后的结果再进行合并), 输出一个参数]
###0 + (0+1+2+3+4 + 0+5+6+7+8+9)
rdd1.aggregate(0)(_+_, _+_)
rdd1.aggregate(0)(math.max(_, _), _ + _)
###5和1比, 得5再和234比得5 --> 5和6789比,得9 --> 5 + (5+9)
rdd1.aggregate(5)(math.max(_, _), _ + _)
val rdd2 = sc.parallelize(List("a","b","c","d","e","f"),2)
def func2(index: Int, iter: Iterator[(String)]) : Iterator[String] = {
iter.toList.map(x => "[partID:" + index + ", val: " + x + "]").iterator
}
rdd2.aggregate("")(_ + _, _ + _)
rdd2.aggregate("=")(_ + _, _ + _)
val rdd3 = sc.parallelize(List("12","23","345","4567"),2)
rdd3.aggregate("")((x,y) => math.max(x.length, y.length).toString, (x,y) => x + y)
val rdd4 = sc.parallelize(List("12","23","345",""),2)
rdd4.aggregate("")((x,y) => math.min(x.length, y.length).toString, (x,y) => x + y)
val rdd5 = sc.parallelize(List("12","23","","345"),2)
rdd5.aggregate("")((x,y) => math.min(x.length, y.length).toString, (x,y) => x + y)
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aggregateByKey
val pairRDD = sc.parallelize(List( ("cat",2), ("cat", 5), ("mouse", 4),
("cat", 12), ("dog", 12), ("mouse", 2)), 2)
def func2(index: Int, iter: Iterator[(String, Int)]) : Iterator[String] = {
iter.toList.map(x => "[partID:" + index + ", val: " + x + "]").iterator
}
pairRDD.mapPartitionsWithIndex(func2).collect
pairRDD.aggregateByKey(0)(math.max(_, _), _ + _).collect
pairRDD.aggregateByKey(100)(math.max(_, _), _ + _).collect
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checkpoint
sc.setCheckpointDir("hdfs://node-1.itcast.cn:9000/ck")
val rdd = sc.textFile("hdfs://node-1.itcast.cn:9000/wc").flatMap(_.split(" ")).map((_, 1)).reduceByKey(_+_)
rdd.checkpoint
rdd.isCheckpointed
rdd.count
rdd.isCheckpointed
rdd.getCheckpointFile
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coalesce, repartition
val rdd1 = sc.parallelize(1 to 10, 10)
val rdd2 = rdd1.coalesce(2, false)
rdd2.partitions.length
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collectAsMap : Map(b -> 2, a -> 1)
val rdd = sc.parallelize(List(("a", 1), ("b", 2)))
rdd.collectAsMap
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combineByKey : 和reduceByKey是相同的效果
###第一个参数x:原封不动取出来, 第二个参数:是函数, 局部运算, 第三个:是函数, 对局部运算后的结果再做运算
###每个分区中每个key中value中的第一个值, (hello,1)(hello,1)(good,1)-->(hello(1,1),good(1))-->x就相当于hello的第一个1, good中的1
val rdd1 = sc.textFile("hdfs://master:9000/wordcount/input/").flatMap(_.split(" ")).map((_, 1))
val rdd2 = rdd1.combineByKey(x => x, (a: Int, b: Int) => a + b, (m: Int, n: Int) => m + n)
rdd1.collect
rdd2.collect
###当input下有3个文件时(有3个block块, 不是有3个文件就有3个block, ), 每个会多加3个10
val rdd3 = rdd1.combineByKey(x => x + 10, (a: Int, b: Int) => a + b, (m: Int, n: Int) => m + n)
rdd3.collect
val rdd4 = sc.parallelize(List("dog","cat","gnu","salmon","rabbit","turkey","wolf","bear","bee"), 3)
val rdd5 = sc.parallelize(List(1,1,2,2,2,1,2,2,2), 3)
val rdd6 = rdd5.zip(rdd4)
val rdd7 = rdd6.combineByKey(List(_), (x: List[String], y: String) => x :+ y, (m: List[String], n: List[String]) => m ++ n)
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countByKey
val rdd1 = sc.parallelize(List(("a", 1), ("b", 2), ("b", 2), ("c", 2), ("c", 1)))
rdd1.countByKey
rdd1.countByValue
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filterByRange
val rdd1 = sc.parallelize(List(("e", 5), ("c", 3), ("d", 4), ("c", 2), ("a", 1)))
val rdd2 = rdd1.filterByRange("b", "d")
rdd2.collect
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flatMapValues : Array((a,1), (a,2), (b,3), (b,4))
val rdd3 = sc.parallelize(List(("a", "1 2"), ("b", "3 4")))
val rdd4 = rdd3.flatMapValues(_.split(" "))
rdd4.collect
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foldByKey
val rdd1 = sc.parallelize(List("dog", "wolf", "cat", "bear"), 2)
val rdd2 = rdd1.map(x => (x.length, x))
val rdd3 = rdd2.foldByKey("")(_+_)
val rdd = sc.textFile("hdfs://node-1.itcast.cn:9000/wc").flatMap(_.split(" ")).map((_, 1))
rdd.foldByKey(0)(_+_)
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foreachPartition
val rdd1 = sc.parallelize(List(1, 2, 3, 4, 5, 6, 7, 8, 9), 3)
rdd1.foreachPartition(x => println(x.reduce(_ + _)))
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keyBy : 以传入的参数做key
val rdd1 = sc.parallelize(List("dog", "salmon", "salmon", "rat", "elephant"), 3)
val rdd2 = rdd1.keyBy(_.length)
rdd2.collect
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keys values
val rdd1 = sc.parallelize(List("dog", "tiger", "lion", "cat", "panther", "eagle"), 2)
val rdd2 = rdd1.map(x => (x.length, x))
rdd2.keys.collect
rdd2.values.collect
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mapPartitions