spark RDD算子(三) distinct,union,intersection,subtract,cartesian

来源:互联网 发布:计算机中文编程 编辑:程序博客网 时间:2024/06/05 06:59

spark伪集合
尽管 RDD 本身不是严格意义上的集合,但它也支持许多数学上的集合操作,比如合并和相交操作, 下图展示了这四种操作
RDD伪集合

distinct

distinct用于去重, 我们生成的RDD可能有重复的元素,使用distinct方法可以去掉重复的元素, 不过此方法涉及到混洗,操作开销很大
scala版本

    scala> var RDD1 = sc.parallelize(List("aa","aa","bb","cc","dd"))    scala> RDD1.collect    res3: Array[String] = Array(aa, aa, bb, cc, dd)    scala> var distinctRDD = RDD1.distinct    scala> distinctRDD.collect    res5: Array[String] = Array(aa, dd, bb, cc)

java版本

    JavaRDD<String> RDD1 = sc.parallelize(Arrays.asList("aa", "aa", "bb", "cc", "dd"));    JavaRDD<String> distinctRDD = RDD1.distinct();    List<String> collect = distinctRDD.collect();    for (String str:collect) {        System.out.print(str+", ");    }---------输出----------aa, dd, bb, cc,

union

两个RDD进行合并
scala版本

    scala> var RDD1 = sc.parallelize(List("aa","aa","bb","cc","dd"))    scala> var RDD2 = sc.parallelize(List("aa","dd","ff"))    scala> RDD1.collect    res6: Array[String] = Array(aa, aa, bb, cc, dd)    scala> RDD2.collect    res7: Array[String] = Array(aa, dd, ff)    scala> RDD1.union(RDD2).collect    res8: Array[String] = Array(aa, aa, bb, cc, dd, aa, dd, ff)

java版本

    JavaRDD<String> RDD1 = sc.parallelize(Arrays.asList("aa", "aa", "bb", "cc", "dd"));    JavaRDD<String> RDD2 = sc.parallelize(Arrays.asList("aa","dd","ff"));    JavaRDD<String> unionRDD = RDD1.union(RDD2);    List<String> collect = unionRDD.collect();    for (String str:collect) {        System.out.print(str+", ");    }-----------输出---------aa, aa, bb, cc, dd, aa, dd, ff,

intersection

RDD1.intersection(RDD2) 返回两个RDD的交集,并且去重
intersection 需要混洗数据,比较浪费性能
scala版本

    scala> var RDD1 = sc.parallelize(List("aa","aa","bb","cc","dd"))    scala> var RDD2 = sc.parallelize(List("aa","dd","ff"))    scala> RDD1.collect    res6: Array[String] = Array(aa, aa, bb, cc, dd)    scala> RDD2.collect    res7: Array[String] = Array(aa, dd, ff)    scala> var insertsectionRDD = RDD1.intersection(RDD2)    scala> insertsectionRDD.collect    res9: Array[String] = Array(aa, dd)

java版本

    JavaRDD<String> RDD1 = sc.parallelize(Arrays.asList("aa", "aa", "bb", "cc", "dd"));    JavaRDD<String> RDD2 = sc.parallelize(Arrays.asList("aa","dd","ff"));    JavaRDD<String> intersectionRDD = RDD1.intersection(RDD2);    List<String> collect = intersectionRDD.collect();    for (String str:collect) {        System.out.print(str+" ");    }-------------输出-----------aa dd

subtract

RDD1.subtract(RDD2),返回在RDD1中出现,但是不在RDD2中出现的元素,不去重
scala版本

    JavaRDD<String> RDD1 = sc.parallelize(Arrays.asList("aa", "aa","bb", "cc", "dd"));    JavaRDD<String> RDD2 = sc.parallelize(Arrays.asList("aa","dd","ff"));    scala> var substractRDD =RDD1.subtract(RDD2)    scala>  substractRDD.collect    res10: Array[String] = Array(bb, cc)

java版本

    JavaRDD<String> RDD1 = sc.parallelize(Arrays.asList("aa", "aa", "bb","cc", "dd"));    JavaRDD<String> RDD2 = sc.parallelize(Arrays.asList("aa","dd","ff"));    JavaRDD<String> subtractRDD = RDD1.subtract(RDD2);    List<String> collect = subtractRDD.collect();    for (String str:collect) {        System.out.print(str+" ");    }------------输出-----------------bb  cc 

cartesian

RDD1.cartesian(RDD2) 返回RDD1和RDD2的笛卡儿积,这个开销非常大

scala版本

    scala>  var RDD1 = sc.parallelize(List("1","2","3"))    scala> var RDD2 = sc.parallelize(List("a","b","c"))    scala> var cartesianRDD = RDD1.cartesian(RDD2)    scala> cartesianRDD.collect    res11: Array[(String, String)] = Array((1,a), (1,b), (1,c), (2,a), (2,b), (2,c), (3,a), (3,b), (3,c))

java版本

    JavaRDD<String> RDD1 = sc.parallelize(Arrays.asList("1", "2", "3"));    JavaRDD<String> RDD2 = sc.parallelize(Arrays.asList("a","b","c"));    JavaPairRDD<String, String> cartesian = RDD1.cartesian(RDD2);    List<Tuple2<String, String>> collect1 = cartesian.collect();    for (Tuple2<String, String> tp:collect1) {        System.out.println("("+tp._1+" "+tp._2+")");    }------------输出-----------------(1 a)(1 b)(1 c)(2 a)(2 b)(2 c)(3 a)(3 b)(3 c)
1 0
原创粉丝点击