Apache spark Spark SQL-重命名列会影响分区吗?
我编写了一个显式连接API,它使用l#或r#前缀重命名数据集中的列,以消除歧义并解决spark沿袭的问题,即columnName1#77在columnName1#123、columnName2#55中找不到 部分代码如下所示:Apache spark Spark SQL-重命名列会影响分区吗?,apache-spark,apache-spark-sql,Apache Spark,Apache Spark Sql,我编写了一个显式连接API,它使用l#或r#前缀重命名数据集中的列,以消除歧义并解决spark沿袭的问题,即columnName1#77在columnName1#123、columnName2#55中找不到 部分代码如下所示: def explicitJoin(other: Dataset[_], joinExpr: Column, joinType: String): ExplicitJoinExt = { val left = dataset.toDF(dataset.columns.
def explicitJoin(other: Dataset[_], joinExpr: Column, joinType: String): ExplicitJoinExt = {
val left = dataset.toDF(dataset.columns.map("l_" + _): _*)
val right = other.toDF(other.columns.map("r_" + _): _*)
new ExplicitJoinExt(left.join(right, joinExpr, joinType))
}
然后,用户可以传递联接表达式,如$“l\U columnName1”===$“r\U columnName1”&&。。。这样他们就可以100%明确地知道他们在连接哪些列
我遇到了一个新问题,分区太大,无法加载到内存中(org.apache.spark.shuffle.FetchFailedException:too large frame…),但是读取输入(分区)数据集没有问题
重命名列是否会影响输入数据集/数据帧的基本分区
编辑
示例1-常规联接
case class A(a: Int, b: String)
val l = (0 to 1000000).map(i => A(i, i.toString))
val r = (0 to 1000000).map(i => A(i, i.toString))
val ds1 = l.toDF.as[A].repartition(100, $"a")
val ds2 = r.toDF.as[A].repartition(100, $"a")
val joined = ds1.join(ds2, Seq("a"), "inner")
joined.explain
== Physical Plan ==
*Project [a#2, b#3, b#15]
+- *SortMergeJoin [a#2], [a#14], Inner
:- *Sort [a#2 ASC NULLS FIRST], false, 0
: +- Exchange hashpartitioning(a#2, 100)
: +- LocalTableScan [a#2, b#3]
+- *Sort [a#14 ASC NULLS FIRST], false, 0
+- ReusedExchange [a#14, b#15], Exchange hashpartitioning(a#2, 100)
示例2-使用涉及重命名的my(可能被误导的)ExplicitJoinText
val joined = ds1
.explicitJoin(ds2, $"l_a" === $"r_a", "inner") // Pimped on conversion to ExplicitJoin type, columns prefixed by l_ or r_. DS joined by expr and join type
.selectLeft // Select just left prefixed columns
.toDF // Convert back from ExplicitJoinExpr to DF
.as[A]
joined.explain
== Physical Plan ==
*Project [l_a#24 AS a#53, l_b#25 AS b#54]
+- *BroadcastHashJoin [l_a#24], [r_a#29], Inner, BuildRight
:- *Project [a#2 AS l_a#24, b#3 AS l_b#25]
: +- Exchange hashpartitioning(a#2, 100)
: +- LocalTableScan [a#2, b#3]
+- BroadcastExchange HashedRelationBroadcastMode(List(cast(input[0, int, false] as bigint)))
+- *Project [a#14 AS r_a#29]
+- Exchange hashpartitioning(a#14, 100)
+- LocalTableScan [a#14]
因此,对于第二个连接,我们将再次重新分区-正确吗?否,我检查了SPARK 2.3.1。重命名不会影响分区,至少在这种方法中不会:
val ds11 = ds1.repartition(4)
val ds11 = ds1.repartition(2, $"cityid")
不,我也查过了。重命名不会影响分区,至少在这种方法中不会:
val ds11 = ds1.repartition(4)
val ds11 = ds1.repartition(2, $"cityid")
解释以下各项的输出:
val j = left.join(right, $"l_personid" === $"r_personid", "inner").explain
在我的例子中,将2和4显示为分区数:
== Physical Plan ==
*(2) BroadcastHashJoin [l_personid#641], [r_personid#647], Inner,
BuildRight, false
:- *(2) Project [personid#612 AS l_personid#641, personname#613 AS
l_personname#642, cityid#614 AS l_cityid#643]
: +- Exchange hashpartitioning(cityid#614, 2)
: +- LocalTableScan [personid#612, personname#613, cityid#614]
+- BroadcastExchange HashedRelationBroadcastMode(List(cast(input[0, int, false] as bigint)))
+- *(1) Project [personid#612 AS r_personid#647, personname#613 AS r_personname#648, cityid#614 AS r_cityid#649]
+- Exchange hashpartitioning(personid#612, 4)
+- LocalTableScan [personid#612, personname#613, cityid#614]
可以看到重命名的col被映射回它们的原始名称
在对其他地方的一篇文章的测试中,我们能够确定依赖聚合或联接的新操作将默认为200,除非
sqlContext.setConf("spark.sql.shuffle.partitions", "some val")
在将其设置为所需值的代码中发出。如果是一小部分数据被合并,等等,那么结果可能会有所不同。对于仍然遇到此问题的人:重命名列确实会影响Spark<3.0中的分区
Seq((1, 2))
.toDF("a", "b")
.repartition($"b")
.withColumnRenamed("b", "c")
.repartition($"c")
.explain()
给出了以下计划:
== Physical Plan ==
Exchange hashpartitioning(c#40, 10)
+- *(1) Project [a#36, b#37 AS c#40]
+- Exchange hashpartitioning(b#37, 10)
+- LocalTableScan [a#36, b#37]
这是固定的。我担心的是,在连接之前,我在(比如)两个DS上对joinKey1进行分区,但在连接$“l_joinKey1”==$“r_joinKey1”之前,我重命名为l_joinKey1和r_joinKey1”-Spark是否足够聪明,能够意识到数据已经在原始列名上进行了分区和共定位?我检查了场景(在午餐期间)并发现分区是按数字继承的,而不是按默认值200继承的。我推断没有问题。你试过解释吗?关于更改名称之前和之后?这个问题:重命名列是否会影响输入数据集/数据帧的基本划分?我相信已经得到了回答。如解释中所示,重命名的列映射回原始列。我最初应用的分区仍然有效。