假设我有这样的数据帧。
val df = sc.parallelize(Seq(
(1.0, 1,"Matt"),
(1.0, 2,"John"),
(1.0, 3,null.asInstanceOf[String]),
(-1.0, 2,"Adam"),
(-1.0, 4,"Steve"))
).toDF("id", "timestamp","name")
我想获取按时间戳排序的每个id的最后一个非null值。这是我的窗口
val partitionWindow = Window.partitionBy($"id").orderBy($"timestamp".desc)
我正在创建一个独特的窗口数据
val filteredDF = df.filter($"name".isNotNull).withColumn("firstName", first("name") over (partitionWindow)).drop("timestamp","name").distinct
并将其加回实际数据
val joinedDF = df.join(filteredDF, windowDF.col("id") === filteredDF.col("id")).drop(filteredDF.col("id"))
joinedDF.show()
它工作正常,但我不喜欢这个解决方案,有人能建议我更好的东西吗?
此外,谁能告诉我为什么最后一个功能不起作用?我试过这个并且结果不正确
val partitionWindow = Window.partitionBy($"id").orderBy($"timestamp")
val windowDF = df.withColumn("lastName", last("name") over (partitionWindow))
答案 0 :(得分:3)
如果要传播最后一个已知值(它与join
使用的逻辑不同),您应该:
ORDER BY timestamp
。last
忽略nulls
:val partitionWindow = Window.partitionBy($"id").orderBy($"timestamp")
df.withColumn("lastName", last("name", true) over (partitionWindow)).show
// +----+---------+-----+--------+
// | id|timestamp| name|lastName|
// +----+---------+-----+--------+
// |-1.0| 2| Adam| Adam|
// |-1.0| 4|Steve| Steve|
// | 1.0| 1| Matt| Matt|
// | 1.0| 2| John| John|
// | 1.0| 3| null| John|
// +----+---------+-----+--------+
如果您想全局采用最后一个值:
ORDER BY timestamp
。last
忽略nulls
:val partitionWindow = Window.partitionBy($"id").orderBy($"timestamp")
.rowsBetween(Window.unboundedPreceding, Window.unboundedFollowing)
df.withColumn("lastName", last("name", true) over (partitionWindow)).show
// +----+---------+-----+--------+
// | id|timestamp| name|lastName|
// +----+---------+-----+--------+
// |-1.0| 2| Adam| Steve|
// |-1.0| 4|Steve| Steve|
// | 1.0| 1| Matt| John|
// | 1.0| 2| John| John|
// | 1.0| 3| null| John|
// +----+---------+-----+--------+