[英]Spark Scala update dataframe
我有这样的问题:
val data = Seq(("TIM", "FIRST", "A", 1),
("BIM", "SECOND", "A", 2),
("JIM", "THIRD", "B", 1)).toDF("NAME", "POSITION", "GROUP", "INDEX")
data.show()
data.printSchema()
val title = Seq(("A", "MASTER"), ("B", "TEACHER"),
("C", "STUDENT")).toDF("LETTER", "DEGREE")
title.show()
title.printSchema()
+----+--------+-----+-----+
|NAME|POSITION|GROUP|INDEX|
+----+--------+-----+-----+
| TIM| FIRST| A| 1|
| BIM| SECOND| A| 2|
| JIM| THIRD| B| 1|
+----+--------+-----+-----+
root
|-- NAME: string (nullable = true)
|-- POSITION: string (nullable = true)
|-- GROUP: string (nullable = true)
|-- INDEX: integer (nullable = false)
+------+-------+
|LETTER| DEGREE|
+------+-------+
| A| MASTER|
| B|TEACHER|
| C|STUDENT|
+------+-------+
root
|-- LETTER: string (nullable = true)
|-- DEGREE: string (nullable = true)
//Final result
+----+--------+-------+--'--+
|NAME|POSITION| GROUP|INDEX|
+----+--------+-------+-----+
| TIM| FIRST| MASTER| 1 |
| BIM| SECOND| A| 2 |
| JIM| THIRD|TEACHER| 1 |
+----+--------+-------+-----+
我尝试了几件事:
val result = data.withColumn("GROUP", when('INDEX === 1, ???????????))
问号在哪里我尝试调用 UDF 但我无法从 GROUP 获取当前行值作为参数传递给 UDF。 还尝试将 select 放在 TITLE 和 GROUP = LETTER 中,但没有任何效果。
首先 dataframe 很大,其他的产量很小。
是否有一些优雅的方式没有先加入它们然后加入 withColumn ?
谢谢
使用广播连接:
data
.join(broadcast(title),$"GROUP"===$"LETTER")
.withColumn("GROUP",when($"INDEX"=== 1,$"DEGREE").otherwise($"GROUP"))
.drop("LETTER","DEGREE")
.show()
+----+--------+-------+-----+
|NAME|POSITION| GROUP|INDEX|
+----+--------+-------+-----+
| TIM| FIRST| MASTER| 1|
| BIM| SECOND| A| 2|
| JIM| THIRD|TEACHER| 1|
+----+--------+-------+-----+
您还可以收集查找地图的title
,广播此 map 并使用 UDF,但与广播连接相比确实没有优势
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.