Question

我有一个如下所示的数据框，并希望通过合并相邻的行来减少它们，即previous.close = current.open

val df = Seq(
  ("Ray","2018-09-01","2018-09-10"),
  ("Ray","2018-09-10","2018-09-15"),
  ("Ray","2018-09-16","2018-09-18"),
  ("Ray","2018-09-21","2018-09-27"),
  ("Ray","2018-09-27","2018-09-30"),
  ("Scott","2018-09-21","2018-09-23"),
  ("Scott","2018-09-24","2018-09-28"),
  ("Scott","2018-09-28","2018-09-30"),
  ("Scott","2018-10-05","2018-10-09"),
  ("Scott","2018-10-11","2018-10-15"),
  ("Scott","2018-10-15","2018-09-20")
)

所需的输出如下：

  (("Ray","2018-09-01","2018-09-15"),
  ("Ray","2018-09-16","2018-09-18"),
  ("Ray","2018-09-21","2018-09-30"),
  ("Scott","2018-09-21","2018-09-23"),
  ("Scott","2018-09-24","2018-09-30"),
  ("Scott","2018-10-05","2018-10-09"),
  ("Scott","2018-10-11","2018-10-20"))

到目前为止，我可以使用下面的DF（）解决方案来压缩相邻行。

df.alias("t1").join(df.alias("t2"),$"t1.name" === $"t2.name" and $"t1.close"=== $"t2.open" )
  .select("t1.name","t1.open","t2.close")
  .distinct.show(false) 

|name |open      |close     |
+-----+----------+----------+
|Scott|2018-09-24|2018-09-30|
|Scott|2018-10-11|2018-09-20|
|Ray  |2018-09-01|2018-09-15|
|Ray  |2018-09-21|2018-09-30|
+-----+----------+----------+

我试图通过给$“ t1.close” =！= $“ t2.open”使用相似的样式来获取单行，然后将两者合并以得到最终结果。但是我得到了多余的行，但是我无法正确过滤。如何实现这一目标？

此帖子与Spark SQL window function with complex condition不同，该帖子将其他日期列计算为新列。

Answer 1

这是一种方法：

如果当前temp1等于先前的null，则用open值创建新列close；否则为当前open
创建另一列temp2，该列用null非空值回填temp1中的last
通过（name，temp2）对结果数据集进行分组以生成连续的日期范围

我已经修改了您的示例数据，以涵盖连续2天以上的连续日期范围的情况。

import org.apache.spark.sql.functions._
import org.apache.spark.sql.expressions.Window

val df = Seq(
  ("Ray","2018-09-01","2018-09-10"),
  ("Ray","2018-09-10","2018-09-15"),
  ("Ray","2018-09-16","2018-09-18"),
  ("Ray","2018-09-21","2018-09-27"),
  ("Ray","2018-09-27","2018-09-30"),
  ("Scott","2018-09-21","2018-09-23"),
  ("Scott","2018-09-23","2018-09-28"),  // <-- Revised
  ("Scott","2018-09-28","2018-09-30"),
  ("Scott","2018-10-05","2018-10-09"),
  ("Scott","2018-10-11","2018-10-15"),
  ("Scott","2018-10-15","2018-10-20")
).toDF("name", "open", "close")

val win = Window.partitionBy($"name").orderBy("open", "close")

val df2 = df.
  withColumn("temp1", when(
    row_number.over(win) === 1 || lag($"close", 1).over(win) =!= $"open", $"open")
  ).
  withColumn("temp2", last($"temp1", ignoreNulls=true).over(
    win.rowsBetween(Window.unboundedPreceding, 0)
  ))

df2.show
// +-----+----------+----------+----------+----------+
// | name|      open|     close|     temp1|     temp2|
// +-----+----------+----------+----------+----------+
// |Scott|2018-09-21|2018-09-23|2018-09-21|2018-09-21|
// |Scott|2018-09-23|2018-09-28|      null|2018-09-21|
// |Scott|2018-09-28|2018-09-30|      null|2018-09-21|
// |Scott|2018-10-05|2018-10-09|2018-10-05|2018-10-05|
// |Scott|2018-10-11|2018-10-15|2018-10-11|2018-10-11|
// |Scott|2018-10-15|2018-10-20|      null|2018-10-11|
// |  Ray|2018-09-01|2018-09-10|2018-09-01|2018-09-01|
// |  Ray|2018-09-10|2018-09-15|      null|2018-09-01|
// |  Ray|2018-09-16|2018-09-18|2018-09-16|2018-09-16|
// |  Ray|2018-09-21|2018-09-27|2018-09-21|2018-09-21|
// |  Ray|2018-09-27|2018-09-30|      null|2018-09-21|
// +-----+----------+----------+----------+----------+

上面显示了步骤1和2的结果，其中temp2保留了相应连续日期范围中最早的open的值。步骤3使用max获取日期范围的最新close：

df2.
  groupBy($"name", $"temp2".as("open")).agg(max($"close").as("close")).
  show
// +-----+----------+----------+
// |name |open      |close     |
// +-----+----------+----------+
// |Scott|2018-09-21|2018-09-30|
// |Scott|2018-10-05|2018-10-09|
// |Scott|2018-10-11|2018-10-20|
// |Ray  |2018-09-01|2018-09-15|
// |Ray  |2018-09-16|2018-09-18|
// |Ray  |2018-09-21|2018-09-30|
// +-----+----------+----------+

Answer 2

已更新：代码现已通过测试：-）

import org.apache.spark.sql.expressions.Window
import org.apache.spark.sql.functions.{coalesce, datediff, lag, lit, min, sum}

val df = Seq(
  ("Ray","2018-09-01","2018-09-10"),
  ("Ray","2018-09-10","2018-09-15"),
  ("Ray","2018-09-16","2018-09-18"),
  ("Ray","2018-09-21","2018-09-27"),
  ("Ray","2018-09-27","2018-09-30"),
  ("Scott","2018-09-21","2018-09-23"),
  ("Scott","2018-09-23","2018-09-28"),  // <-- Revised
  ("Scott","2018-09-28","2018-09-30"),
  ("Scott","2018-10-05","2018-10-09"),
  ("Scott","2018-10-11","2018-10-15"),
  ("Scott","2018-10-15","2018-10-20")
).toDF("name", "open", "close")

val window = Window.partitionBy("name").orderBy($"open").rowsBetween(-1, Window.currentRow) //<- only compare the dates of a certain name, and for each row look also look at the previous one

df.select(
  $"name", $"open", $"close",
  min($"close").over(window) as "closeBefore_tmp"//<- get the smaller close value (that of the previous entry) 
)
.withColumn("closeBefore", when($"closeBefore_tmp" === $"close", null).otherwise($"closeBefore_tmp")) //<- in this case there was no previous row: its the first for this user, so set closeBefore to null
.createOrReplaceTempView("tmp")

现在您可以compare打开并closeBefore。

根据相邻的行和单个行来产生火花并减少火花

2 个答案: