scala - 使用 DataFrame API 时，自联接无法按预期工作

Question

我正在尝试使用自联接从表中获取最新记录。它使用 spark-sql但无法使用 spark DataFrame API。

任何人都可以帮忙吗？它是一个错误吗？

我在本地模式下使用 Spark 2.2.0

创建输入 DataFrame :

scala> val df3 = spark.sparkContext.parallelize(Array((1,"a",1),(1,"aa",2),(2,"b",2),(2,"bb",5))).toDF("id","value","time")
df3: org.apache.spark.sql.DataFrame = [id: int, value: string ... 1 more field]    

scala> val df33 = df3
df33: org.apache.spark.sql.DataFrame = [id: int, value: string ... 1 more field]

scala> df3.show
+---+-----+----+
| id|value|time|
+---+-----+----+
|  1|    a|   1|
|  1|   aa|   2|
|  2|    b|   2|
|  2|   bb|   5|
+---+-----+----+

scala> df33.show
+---+-----+----+
| id|value|time|
+---+-----+----+
|  1|    a|   1|
|  1|   aa|   2|
|  2|    b|   2|
|  2|   bb|   5|
+---+-----+----+

现在使用 SQL 执行连接: 作品

scala> spark.sql("select df33.* from df3 join df33 on df3.id = df33.id and df3.time < df33.time").show
+---+-----+----+
| id|value|time|
+---+-----+----+
|  1|   aa|   2|
|  2|   bb|   5|
+---+-----+----+

现在使用数据框 API 执行连接: 不起作用

scala> df3.join(df33, (df3.col("id") === df33.col("id")) && (df3.col("time") < df33.col("time")) ).select(df33.col("id"),df33.col("value"),df33.col("time")).show
+---+-----+----+
| id|value|time|
+---+-----+----+
+---+-----+----+

需要注意的是解释计划: DataFrame 为空白API!!

scala> df3.join(df33, (df3.col("id") === df33.col("id")) && (df3.col("time") < df33.col("time")) ).select(df33.col("id"),df33.col("value"),df33.col("time")).explain
== Physical Plan ==
LocalTableScan <empty>, [id#150, value#151, time#152]

scala> spark.sql("select df33.* from df3 join df33 on df3.id = df33.id and df3.time < df33.time").explain
== Physical Plan ==
*Project [id#1241, value#1242, time#1243]
+- *SortMergeJoin [id#150], [id#1241], Inner, (time#152 < time#1243)
   :- *Sort [id#150 ASC NULLS FIRST], false, 0
   :  +- Exchange hashpartitioning(id#150, 200)
   :     +- *Project [_1#146 AS id#150, _3#148 AS time#152]
   :        +- *SerializeFromObject [assertnotnull(input[0, scala.Tuple3, true])._1 AS _1#146, staticinvoke(class org.apache.spark.unsafe.types.UTF8String, StringType, fromString,
assertnotnull(input[0, scala.Tuple3, true])._2, true) AS _2#147, assertnotnull(input[0, scala.Tuple3, true])._3 AS _3#148]
   :           +- Scan ExternalRDDScan[obj#145]
   +- *Sort [id#1241 ASC NULLS FIRST], false, 0
      +- Exchange hashpartitioning(id#1241, 200)
         +- *Project [_1#146 AS id#1241, _2#147 AS value#1242, _3#148 AS time#1243]
            +- *SerializeFromObject [assertnotnull(input[0, scala.Tuple3, true])._1 AS _1#146, staticinvoke(class org.apache.spark.unsafe.types.UTF8String, StringType, fromString,
assertnotnull(input[0, scala.Tuple3, true])._2, true) AS _2#147, assertnotnull(input[0, scala.Tuple3, true])._3 AS _3#148]
               +- Scan ExternalRDDScan[obj#145]

Answer 1

不，这不是一个错误，但是当您像您所做的那样将 DataFrame 重新分配给一个新的时，它实际上会复制沿袭但不会复制数据。因此，您将在同一列上进行比较。

使用spark.sql略有不同，因为它实际上是在处理您的 DataFrame 的别名。 s

因此，使用 API 执行自联接的正确方法实际上是为您的 DataFrame 设置别名。如下:

val df1 = Seq((1,"a",1),(1,"aa",2),(2,"b",2),(2,"bb",5)).toDF("id","value","time")

df1.as("df1").join(df1.as("df2"), $"df1.id" === $"df2.id" && $"df1.time" < $"df2.time").select($"df2.*").show
// +---+-----+----+
// | id|value|time|
// +---+-----+----+
// |  1|   aa|   2|
// |  2|   bb|   5|
// +---+-----+----+

有关自联接的更多信息，我建议阅读 High Performance Spark by Rachel Warren, Holden Karau - Chapter 4 .