Get the row corresponding to the latest timestamp in a Spark Dataset using Scala

8
votes

I am relatively new to Spark and Scala. I have a dataframe which has the following format:

| Col1 | Col2 | Col3 | Col_4 | Col_5 | Col_TS                  | Col_7 | 

| 1234 | AAAA | 1111 | afsdf | ewqre | 1970-01-01 00:00:00.0   | false |
| 1234 | AAAA | 1111 | ewqrw | dafda | 2017-01-17 07:09:32.748 | true  |
| 1234 | AAAA | 1111 | dafsd | afwew | 2015-01-17 07:09:32.748 | false |
| 5678 | BBBB | 2222 | afsdf | qwerq | 1970-01-01 00:00:00.0   | true  |
| 5678 | BBBB | 2222 | bafva | qweqe | 2016-12-08 07:58:43.04  | false |
| 9101 | CCCC | 3333 | caxad | fsdaa | 1970-01-01 00:00:00.0   | false |

What I need to do is to get the row that corresponds to the latest timestamp. In the example above, the keys are Col1, Col2 and Col3. Col_TS represents the timestamp and Col_7 is a boolean that determines the validity of the record. What I want to do is to find a way to group these records based on the keys and retain the one that has the latest timestamp.

So the output of the operation in the dataframe above should be:

| Col1 | Col2 | Col3 | Col_4 | Col_5 | Col_TS                  | Col_7 | 

| 1234 | AAAA | 1111 | ewqrw | dafda | 2017-01-17 07:09:32.748 | true  |
| 5678 | BBBB | 2222 | bafva | qweqe | 2016-12-08 07:58:43.04  | false |
| 9101 | CCCC | 3333 | caxad | fsdaa | 1970-01-01 00:00:00.0   | false |

I came up with a partial solution but this way I can only return the dataframe of the Column keys on which the records are grouped and not the other columns. 

df = df.groupBy("Col1","Col2","Col3").agg(max("Col_TS"))


| Col1 | Col2 | Col3 | max(Col_TS)             |

| 1234 | AAAA | 1111 | 2017-01-17 07:09:32.748 |
| 5678 | BBBB | 2222 | 2016-12-08 07:58:43.04  | 
| 9101 | CCCC | 3333 | 1970-01-01 00:00:00.0   |

Can someone help me in coming up with a Scala code for performing this operation?

2
scalaapache-sparkspark-dataframe

2 Answers

16
votes

You can use window function as following 

import org.apache.spark.sql.functions._
val windowSpec = Window.partitionBy("Col1","Col2","Col3").orderBy(col("Col_TS").desc)

df.withColumn("maxTS", first("Col_TS").over(windowSpec))
.select("*").where(col("maxTS") === col("Col_TS"))
.drop("maxTS")
  .show(false)

You should get output as following

+----+----+----+-----+-----+----------------------+-----+
|Col1|Col2|Col3|Col_4|Col_5|Col_TS                |Col_7|
+----+----+----+-----+-----+----------------------+-----+
|5678|BBBB|2222|bafva|qweqe|2016-12-0807:58:43.04 |false|
|1234|AAAA|1111|ewqrw|dafda|2017-01-1707:09:32.748|true |
|9101|CCCC|3333|caxad|fsdaa|1970-01-0100:00:00.0  |false|
+----+----+----+-----+-----+----------------------+-----+
2
votes

One option is firstly order the data frame by Col_TS, then group by Col1, Col2 and Col3 and take the last item from each other column:

val val_columns = Seq("Col_4", "Col_5", "Col_TS", "Col_7").map(x => last(col(x)).alias(x))

(df.orderBy("Col_TS")
   .groupBy("Col1", "Col2", "Col3")
   .agg(val_columns.head, val_columns.tail: _*).show)

+----+----+----+-----+-----+--------------------+-----+
|Col1|Col2|Col3|Col_4|Col_5|              Col_TS|Col_7|
+----+----+----+-----+-----+--------------------+-----+
|1234|AAAA|1111|ewqrw|dafda|2017-01-17 07:09:...| true|
|9101|CCCC|3333|caxad|fsdaa|1970-01-01 00:00:...|false|
|5678|BBBB|2222|bafva|qweqe|2016-12-08 07:58:...|false|
+----+----+----+-----+-----+--------------------+-----+