Reputation: 465
I have a dataframe like this
df.show(5)
kv |list1 |list2 |p
[k1,v2|[1,2,5,9 |[5,1,7,9,6,3,1,4,9] |0.5
[k1,v3|[1,2,5,8,9|[5,1,7,9,6,3,1,4,15] |0.9
[k2,v2|[77,2,5,9]|[0,1,8,9,7,3,1,4,100]|0.01
[k5,v5|[1,0,5,9 |[5,1,7,9,6,3,1,4,3] |0.3
[k9,v2|[1,2,5,9 |[5,1,7,9,6,3,1,4,200]|2.5
df.count()
5200158
I want to get the row that have maximum p, this below works for me but I don't know if there is another cleaner way
val f = df.select(max(struct(
col("pp") +: df.columns.collect { case x if x != "p" => col(x) }: _*
))).first()
Upvotes: 3
Views: 13564
Reputation: 1
Just order by and then take:
import org.apache.spark.sql.functions.desc
df.orderBy(desc("pp")).take(1)
or
df.orderBy(desc("pp")).limit(1).first
Upvotes: 8
Reputation: 27383
You can also use Window-Functions, this is especially useful if the logic of selecting the row gets more complex (other than global min/max) :
import org.apache.spark.sql.expressions.Window
df
.withColumn("max_p",max($"p").over(Window.partitionBy()))
.where($"p" === $"max_p")
.drop($"max_p")
.first()
Upvotes: 4