给定以下数据框架:
+----+-----+---+-----+
| uid| k| v|count|
+----+-----+---+-----+
| a|pref1| b| 168|
| a|pref3| h| 168|
| a|pref3| t| 63|
| a|pref3| k| 84|
| a|pref1| e| 84|
| a|pref2| z| 105|
+----+-----+---+-----+
如何从uid
,k
获得最大值,但包括v
?
+----+-----+---+----------+
| uid| k| v|max(count)|
+----+-----+---+----------+
| a|pref1| b| 168|
| a|pref3| h| 168|
| a|pref2| z| 105|
+----+-----+---+----------+
我可以做这样的事情,但它会删除" V"列:
df.groupBy("uid", "k").max("count")
它是窗口操作员(使用over
函数)或join
的完美示例。
由于您已经弄清楚了如何使用Windows,因此我专注于join
。
scala> val inventory = Seq(
| ("a", "pref1", "b", 168),
| ("a", "pref3", "h", 168),
| ("a", "pref3", "t", 63)).toDF("uid", "k", "v", "count")
inventory: org.apache.spark.sql.DataFrame = [uid: string, k: string ... 2 more fields]
scala> val maxCount = inventory.groupBy("uid", "k").max("count")
maxCount: org.apache.spark.sql.DataFrame = [uid: string, k: string ... 1 more field]
scala> maxCount.show
+---+-----+----------+
|uid| k|max(count)|
+---+-----+----------+
| a|pref3| 168|
| a|pref1| 168|
+---+-----+----------+
scala> val maxCount = inventory.groupBy("uid", "k").agg(max("count") as "max")
maxCount: org.apache.spark.sql.DataFrame = [uid: string, k: string ... 1 more field]
scala> maxCount.show
+---+-----+---+
|uid| k|max|
+---+-----+---+
| a|pref3|168|
| a|pref1|168|
+---+-----+---+
scala> maxCount.join(inventory, Seq("uid", "k")).where($"max" === $"count").show
+---+-----+---+---+-----+
|uid| k|max| v|count|
+---+-----+---+---+-----+
| a|pref3|168| h| 168|
| a|pref1|168| b| 168|
+---+-----+---+---+-----+
这是我到目前为止提出的最好的解决方案:
val w = Window.partitionBy("uid","k").orderBy(col("count").desc)
df.withColumn("rank", dense_rank().over(w)).select("uid", "k","v","count").where("rank == 1").show
您可以使用窗口函数:
from pyspark.sql.functions import max as max_
from pyspark.sql.window import Window
w = Window.partitionBy("uid", "k")
df.withColumn("max_count", max_("count").over(w))