rdd

Spark: subtract two DataFrames

由 喜你入骨 提交于 2019-11-26 08:10:47
问题 In Spark version 1.2.0 one could use subtract with 2 SchemRDD s to end up with only the different content from the first one val onlyNewData = todaySchemaRDD.subtract(yesterdaySchemaRDD) onlyNewData contains the rows in todaySchemRDD that do not exist in yesterdaySchemaRDD . How can this be achieved with DataFrames in Spark version 1.3.0 ? 回答1: According to the api docs, doing: dataFrame1.except(dataFrame2) will return a new DataFrame containing rows in dataFrame1 but not in dataframe2. 回答2:

How to read from hbase using spark

由 China☆狼群 提交于 2019-11-26 07:58:26
问题 The below code will read from the hbase, then convert it to json structure and the convert to schemaRDD , But the problem is that I am using List to store the json string then pass to javaRDD, for data of about 100 GB the master will be loaded with data in memory. What is the right way to load the data from hbase then perform manipulation,then convert to JavaRDD. package hbase_reader; import java.io.IOException; import java.io.Serializable; import java.util.ArrayList; import java.util.List;

Default Partitioning Scheme in Spark

由 南楼画角 提交于 2019-11-26 07:47:20
问题 When I execute below command: scala> val rdd = sc.parallelize(List((1,2),(3,4),(3,6)),4).partitionBy(new HashPartitioner(10)).persist() rdd: org.apache.spark.rdd.RDD[(Int, Int)] = ShuffledRDD[10] at partitionBy at <console>:22 scala> rdd.partitions.size res9: Int = 10 scala> rdd.partitioner.isDefined res10: Boolean = true scala> rdd.partitioner.get res11: org.apache.spark.Partitioner = org.apache.spark.HashPartitioner@a It says that there are 10 partitions and partitioning is done using

Stackoverflow due to long RDD Lineage

由 牧云@^-^@ 提交于 2019-11-26 07:35:04
问题 I have thousands of small files in HDFS. Need to process a slightly smaller subset of files (which is again in thousands), fileList contains list of filepaths which need to be processed. // fileList == list of filepaths in HDFS var masterRDD: org.apache.spark.rdd.RDD[(String, String)] = sparkContext.emptyRDD for (i <- 0 to fileList.size() - 1) { val filePath = fileStatus.get(i) val fileRDD = sparkContext.textFile(filePath) val sampleRDD = fileRDD.filter(line => line.startsWith(\"#####\")).map

ubuntu 升级QT版本后，在新机上发布QT程序报错：qt.qpa.plugin: Could not find the Qt platform plugin “xcb” in “”

由 試著忘記壹切 提交于 2019-11-26 07:33:18
本地原本使用的QT版本是5.11，最近更新成了5.13.2；在本地编译、运行一切正常；发布release版本到新的ubuntu上；由于在新机器上没有安装QT，所以发布的时候通过ldd导致出程序的依赖库，把这些依赖库和程序放在同一目录一起拷贝到新机器上，初次运行需要指定库目录，否则会报找不到库文件的错误： 指定程序执行的库文件目录，在终端输入指令： export LD_LIBRARY_PATH='/home/rdd/pp':$LD_LIBRARY_PATH 然后再重新运行程序，结果运行时出现错误： 猜测是运行依赖库不全导致，但是一开始通过ldd导出的依赖库在编译程序的机器上已经是完整的，所以这里应该是内部库的二次依赖导致。启动QT_DEBUG_PLUGINS，查看程序的执行过程，分析是哪部分缺失。也可以通过设置环境变量的方式，不过如果只是调bug，只需要在终端输入，让其一次性显示即可，在终端输入指令： export QT_DEBUG_PLUGINS=1 然后再重新启动发布的程序，发现多了一点打印信息： 提示在程序的执行目录下，企图加载platforms的相关库，以打印相关的库调用信息，但是这里并没有相关打印。是因为当前缺少platforms目录的库文件导致。 从原机器上(装了QT5.13.2)的QT安装目录中，把该目录以及其内部所有文件完整拷贝到新机器上(xftp拷贝)

Reduce a key-value pair into a key-list pair with Apache Spark

由 人走茶凉 提交于 2019-11-26 07:29:10
问题 I am writing a Spark application and want to combine a set of Key-Value pairs (K, V1), (K, V2), ..., (K, Vn) into one Key-Multivalue pair (K, [V1, V2, ..., Vn]) . I feel like I should be able to do this using the reduceByKey function with something of the flavor: My_KMV = My_KV.reduce(lambda a, b: a.append([b])) The error that I get when this occurs is: \'NoneType\' object has no attribue \'append\'. My keys are integers and values V1,...,Vn are tuples. My goal is to create a single pair with

Which operations preserve RDD order?

由 孤街醉人 提交于 2019-11-26 07:25:17
问题 RDD has a meaningful (as opposed to some random order imposed by the storage model) order if it was processed by sortBy(), as explained in this reply. Now, which operations preserve that order? E.g., is it guaranteed that (after a.sortBy() ) a.map(f).zip(a) === a.map(x => (f(x),x)) How about a.filter(f).map(g) === a.map(x => (x,g(x))).filter(f(_._1)).map(_._2) what about a.filter(f).flatMap(g) === a.flatMap(x => g(x).map((x,_))).filter(f(_._1)).map(_._2) Here \"equality\" === is understood as

spark RDD 的map与flatmap区别说明

由 旧时模样 提交于 2019-11-26 06:36:25
HDFS到HDFS过程 看看 map 和 flatmap 的位置 Flatmap 和 map 的定义 map()是将函数用于RDD中的每个元素，将返回值构成新的RDD。 flatmap()是将函数应用于RDD中的每个元素，将返回的迭代器的所有内容构成新的RDD 例子 ： val rdd = sc.parallelize( List ( "coffee panda" , "happy panda" , "happiest panda party" )) 输入 rdd. map (x=>x).collect 结果 res9: Array [ String ] = Array (coffee panda, happy panda, happiest panda party) 输入 rdd.flatMap(x=>x.split( " " )).collect 结果 res8: Array [ String ] = Array (coffee, panda, happy, panda, happiest, panda, party) flatMap说明白就是先map然后再flat，再来看个例子 val rdd1 = sc.parallelize( List ( 1 , 2 , 3 , 3 )) scala> rdd1.map(x=>x+ 1 ).collect res10: Array

Matrix Multiplication in Apache Spark [closed]

由 别等时光非礼了梦想. 提交于 2019-11-26 06:35:47
问题 I am trying to perform matrix multiplication using Apache Spark and Java. I have 2 main questions: How to create RDD that can represent matrix in Apache Spark? How to multiply two such RDDs? 回答1: All depends on the input data and dimensions but generally speaking what you want is not a RDD but one of the distributed data structures from org.apache.spark.mllib.linalg.distributed. At this moment it provides four different implementations of the DistributedMatrix IndexedRowMatrix - can be

Scala中map操作和flatMap操作的区别

由 感情迁移 提交于 2019-11-26 06:33:35
Scala中map操作和flatMap操作的区别 flatMap()和map() 操作对比 map操作 flatMap操作 加深理解 flatMap()和map() flatMap(function)和map(function)操作都是传入的函数对RDD中的每一个元素进行处理的操作。但是不同点在于，map操作是一对一的，即map(function)传入的函数在处理RDD的每一个元素后都产生相对应的结果，输入与输出是 一对一的 ，并由这些一一对应的结果值获得输出RDD；而flatMap(function)传入的函数在处理RDD的每一个元素时，输入与输出是 一对一或一对多的 ，都可能产生一个或多个对应的元素组成的返回值序列的迭代器，输出的RDD是一个包含各个迭代器可访问的所有元素的RDD。 操作对比 map操作 flatMap操作 加深理解 来源： CSDN 作者： 震逗比 链接： https://blog.csdn.net/zzlove1234567890/article/details/88798954
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