为什么Spark认为这是十字 /笛卡尔的加入

Question

我想两次加入数据：

rdd1 = spark.createDataFrame([(1, 'a'), (2, 'b'), (3, 'c')], ['idx', 'val'])
rdd2 = spark.createDataFrame([(1, 2, 1), (1, 3, 0), (2, 3, 1)], ['key1', 'key2', 'val'])
res1 = rdd1.join(rdd2, on=[rdd1['idx'] == rdd2['key1']])
res2 = res1.join(rdd1, on=[res1['key2'] == rdd1['idx']])
res2.show()

然后我有一些错误：

pyspark.sql.utils.analysisexception：u'cartesian Joins可能是 过于昂贵，默认情况下是禁用的。要明确启用它们，请设置spark.sql.crossjoin.enabled = true;'

，但我认为这不是十字架加入

更新：

res2.explain()
== Physical Plan ==
CartesianProduct
:- *SortMergeJoin [idx#0L, idx#0L], [key1#5L, key2#6L], Inner
:  :- *Sort [idx#0L ASC, idx#0L ASC], false, 0
:  :  +- Exchange hashpartitioning(idx#0L, idx#0L, 200)
:  :     +- *Filter isnotnull(idx#0L)
:  :        +- Scan ExistingRDD[idx#0L,val#1]
:  +- *Sort [key1#5L ASC, key2#6L ASC], false, 0
:     +- Exchange hashpartitioning(key1#5L, key2#6L, 200)
:        +- *Filter ((isnotnull(key2#6L) && (key2#6L = key1#5L)) && isnotnull(key1#5L))
:           +- Scan ExistingRDD[key1#5L,key2#6L,val#7L]
+- Scan ExistingRDD[idx#40L,val#41]

Answer 1

发生这种情况，因为您join结构共享相同的谱系，这导致了一个琐碎的条件：

res2.explain()

== Physical Plan ==
org.apache.spark.sql.AnalysisException: Detected cartesian product for INNER join between logical plans
Join Inner, ((idx#204L = key1#209L) && (key2#210L = idx#204L))
:- Filter isnotnull(idx#204L)
:  +- LogicalRDD [idx#204L, val#205]
+- Filter ((isnotnull(key2#210L) && (key2#210L = key1#209L)) && isnotnull(key1#209L))
   +- LogicalRDD [key1#209L, key2#210L, val#211L]
and
LogicalRDD [idx#235L, val#236]
Join condition is missing or trivial.
Use the CROSS JOIN syntax to allow cartesian products between these relations.;

如果这样，您应该使用别名：

from pyspark.sql.functions import col
rdd1 = spark.createDataFrame(...).alias('rdd1')
rdd2 = spark.createDataFrame(...).alias('rdd2')
res1 = rdd1.join(rdd2, col('rdd1.idx') == col('rdd2.key1')).alias('res1')
res1.join(rdd1, on=col('res1.key2') == col('rdd1.idx')).explain()

== Physical Plan ==
*SortMergeJoin [key2#297L], [idx#360L], Inner
:- *Sort [key2#297L ASC NULLS FIRST], false, 0
:  +- Exchange hashpartitioning(key2#297L, 200)
:     +- *SortMergeJoin [idx#290L], [key1#296L], Inner
:        :- *Sort [idx#290L ASC NULLS FIRST], false, 0
:        :  +- Exchange hashpartitioning(idx#290L, 200)
:        :     +- *Filter isnotnull(idx#290L)
:        :        +- Scan ExistingRDD[idx#290L,val#291]
:        +- *Sort [key1#296L ASC NULLS FIRST], false, 0
:           +- Exchange hashpartitioning(key1#296L, 200)
:              +- *Filter (isnotnull(key2#297L) && isnotnull(key1#296L))
:                 +- Scan ExistingRDD[key1#296L,key2#297L,val#298L]
+- *Sort [idx#360L ASC NULLS FIRST], false, 0
   +- Exchange hashpartitioning(idx#360L, 200)
      +- *Filter isnotnull(idx#360L)
         +- Scan ExistingRDD[idx#360L,val#361]

有关详细信息，请参见Spark-6459。

Answer 2

我在第二次加入之前坚持数据框时也很成功。

类似：

res1 = rdd1.join(rdd2, col('rdd1.idx') == col('rdd2.key1')).persist()
res1.join(rdd1, on=col('res1.key2') == col('rdd1.idx'))

Answer 3

坚持不对我有用。

i用dataframes上的别名克服了它

from pyspark.sql.functions import col
df1.alias("buildings").join(df2.alias("managers"), col("managers.distinguishedName") == col("buildings.manager"))