Question

是火花的新手，所以我需要创建一个只有两个元素的from tkinter import * from tkinter.ttk import * root = Tk() root.title("Playground") #notebook _notebook1 = Notebook(root) _tab1 = Frame(_notebook1) _notebook1.add(_tab1, text = "Tester", sticky = "e") #_notebook1.grid(sticky=E+W+N+S) #_tab1.grid(sticky=E+W+N+S) _notebook1.pack(fill=BOTH, expand=1) _startBtn = Button(_tab1, text="start") _startBtn.grid(sticky=E+W) root.grid_columnconfigure(0, weight=1) root.grid_rowconfigure(0, weight=1) # _tab1.grid_columnconfigure(0, weight=1) # _tab1.grid_rowconfigure(0, weight=1) # _notebook1.grid_columnconfigure(0, weight=1) # _notebook1.grid_rowconfigure(0, weight=1) root.mainloop()。

RDD

当我执行groupby键时，输出为Array1 = ((1,1)(1,2)(1,3),(2,1),(2,2),(2,3)

但是我需要输出与键只有2个值对。我不确定如何获得它。

((1,(1,2,3)),(2,(1,2,3))

这些值只能打印一次。应该只有Expected Output = ((1,(1,2)),(1,(1,3)),(1(2,3),(2(1,2)),(2,(1,3)),(2,(2,3)))而不是(1,2) 或像(2,1)而不是(2,3)

谢谢

Answer 1

您可以按以下方式获得所需的结果：

// Prior to doing the `groupBy`, you have an RDD[(Int, Int)], x, containing:
//   (1,1),(1,2),(1,3),(2,1),(2,2),(2,3)
//
// Can simply map values as below. Result is a RDD[(Int, (Int, Int))].
val x: RDD[(Int, Int)] = sc.parallelize(Seq((1,1),(1,2),(1,3),(2,1),(2,2),(2,3))
val y: RDD[(Int, (Int, Int))] = x.map(t => (t._1, t)) // Map first value in pair tuple to the tuple
y.collect // Get result as an array
// res0: Array[(Int, (Int, Int))] = Array((1,(1,1)), (1,(1,2)), (1,(1,3)), (2,(2,1)), (2,(2,2)), (2,(2,3)))

也就是说，结果是一对RDD，它将键（每对的第一个值）与该对（作为 tuple ）相关联。不要使用groupBy，因为在这种情况下，它不会给您想要的东西。

Answer 2

如果我正确理解了您的要求，则可以使用groupByKey和flatMapValues展平分组值的2-combinations，如下所示：

val rdd = sc.parallelize(Seq(
  (1, 1), (1,  2), (1 ,3), (2, 1), (2, 2), (2, 3)
))

rdd.groupByKey.flatMapValues(_.toList.combinations(2)).
  map{ case (k, v) => (k, (v(0), v(1))) }.
  collect
// res1: Array[(Int, (Int, Int))] =
//   Array((1,(1,2)), (1,(1,3)), (1,(2,3)), (2,(1,2)), (2,(1,3)), (2,(2,3)))

使用scala在火花中创建对RDD

2 个答案: