Scala 如何按两个数据帧中的列分组,然后在行之间应用聚合差异函数?
我有两个数据帧,如下所示:Scala 如何按两个数据帧中的列分组,然后在行之间应用聚合差异函数?,scala,apache-spark,apache-spark-sql,spark-dataframe,Scala,Apache Spark,Apache Spark Sql,Spark Dataframe,我有两个数据帧,如下所示: +--------+----------+------+-------------------+ |readerId|locationId|userId| timestamp| +--------+----------+------+-------------------+ | R2| l1| u2|2018-04-12 05:00:00| | R1| l1| u1|2018-04-12 0
+--------+----------+------+-------------------+
|readerId|locationId|userId| timestamp|
+--------+----------+------+-------------------+
| R2| l1| u2|2018-04-12 05:00:00|
| R1| l1| u1|2018-04-12 05:00:00|
| R3| l3| u3|2018-04-12 05:00:00|
+--------+----------+------+-------------------+
+--------+----------+------+-------------------+
|readerId|locationId|userId| timestamp|
+--------+----------+------+-------------------+
| R1| l1| u1|2018-04-12 07:00:00|
| R2| l1| u2|2018-04-12 10:00:00|
| R3| l3| u3|2018-04-12 07:00:00|
+--------+----------+------+-------------------+
我想对readerId
和locationId
进行分组,然后查找分组值的时间戳差异。例如:对于readerIDR1
,locationIDl1
,时间戳差为2小时
我通过连接两个数据帧并使用with column
实现了它
val joinedDf = asKuduDf.join(
asOutToInDf,
col("kdf.locationId") <=> col("outInDf.locationId") &&
(col("kdf.readerId") <=> col("outInDf.readerId")),
"inner")
//Time loged in calculation
val timestampDf = joinedDf.withColumn(
"totalTime",
((unix_timestamp($"outInDf.timestamp") -
unix_timestamp($"kdf.timestamp"))/60).cast("long")
).toDF()
但上述方法的问题在于没有“差异”函数
join
是正确的解决方案。通常情况下,groupby
和聚合不是一个选项,特别是当(readerId
,locationId
)不是唯一标识符时
你可以
unionDf
.groupBy($"readerId", $"locationId")
.agg((max($"timestamp").cast("long") - min($"timestamp").cast(long) / 60).alias("diff"))
但这是一个高度人工的解决方案,与
join
相比没有任何优势。它对一些微妙的数据问题也很敏感。您可以使用union
合并这两个数据帧,在聚合中,您可以根据需要计算差异
val mergedDF = asKuduDf.union(asOutToInDf)
.groupBy($"readerId", $"locationId")
.agg(collect_list($"timestamp").as("time"))
mergedDF.withColumn("dif",
abs(unix_timestamp($"time" (0)) - unix_timestamp($"time" (1))) / 60
)
输出:
+--------+----------+------------------------------------------+-----+
|readerId|locationId|time |dif |
+--------+----------+------------------------------------------+-----+
|R3 |l3 |[2018-04-12 05:00:00, 2018-04-12 07:00:00]|120.0|
|R2 |l1 |[2018-04-12 05:00:00, 2018-04-12 10:00:00]|300.0|
|R1 |l1 |[2018-04-12 05:00:00, 2018-04-12 07:00:00]|120.0|
+--------+----------+------------------------------------------+-----+
希望这有帮助 虽然您正确地认为
join
是正确的解决方案,但我认为您必须假设(readerId,locationId)
对是唯一的,否则整个问题毫无意义,无论是使用聚合还是连接-当每边都有多个值时,差异的含义是什么?
+--------+----------+------------------------------------------+-----+
|readerId|locationId|time |dif |
+--------+----------+------------------------------------------+-----+
|R3 |l3 |[2018-04-12 05:00:00, 2018-04-12 07:00:00]|120.0|
|R2 |l1 |[2018-04-12 05:00:00, 2018-04-12 10:00:00]|300.0|
|R1 |l1 |[2018-04-12 05:00:00, 2018-04-12 07:00:00]|120.0|
+--------+----------+------------------------------------------+-----+