sql – 在Apache Spark Join中包含空值

这是默认的Spark行为。

val numbersDf = Seq(
  ("123"),("456"),(null),("")
).toDF("numbers")

val lettersDf = Seq(
  ("123","abc"),("456","def"),(null,"zzz"),("","hhh")
).toDF("numbers","letters")

val joinedDf = numbersDf.join(lettersDf,Seq("numbers"))

这是joinedDf.show()的输出：

+-------+-------+
|numbers|letters|
+-------+-------+
|    123|    abc|
|    456|    def|
|       |    hhh|
+-------+-------+

这是我想要的输出：

+-------+-------+
|numbers|letters|
+-------+-------+
|    123|    abc|
|    456|    def|
|       |    hhh|
|   null|    zzz|
+-------+-------+

解决方法

numbersDf
  .join(lettersDf,numbersDf("numbers") <=> lettersDf("numbers"))
  .drop(lettersDf("numbers"))

+-------+-------+
|numbers|letters|
+-------+-------+
|    123|    abc|
|    456|    def|
|   null|    zzz|
|       |    hhh|
+-------+-------+

小心不要在Spark 1.5或更早版本中使用它。在Spark 1.6之前，它需要一个笛卡尔积(SPARK-22222 – 快速零安全连接)。

在Spark 2.3.0或更高版本中，您可以在PySpark中使用Column.eqNullSafe：

numbers_df = sc.parallelize([
    ("123",),(None,)
]).toDF(["numbers"])

letters_df = sc.parallelize([
    ("123","hhh")
]).toDF(["numbers","letters"])

numbers_df.join(letters_df,numbers_df.numbers.eqNullSafe(letters_df.numbers))

+-------+-------+-------+
|numbers|numbers|letters|
+-------+-------+-------+
|    456|    456|    def|
|   null|   null|    zzz|
|       |       |    hhh|
|    123|    123|    abc|
+-------+-------+-------+

和SparkR中的％< =>％：

numbers_df <- createDataFrame(data.frame(numbers = c("123","456",NA,"")))
letters_df <- createDataFrame(data.frame(
  numbers = c("123",""),letters = c("abc","def","zzz","hhh")
))

head(join(numbers_df,letters_df,numbers_df$numbers %<=>% letters_df$numbers))

numbers numbers letters
1     456     456     def
2    <NA>    <NA>     zzz
3                     hhh
4     123     123     abc

使用SQL(Spark 2.2.0)，您可以使用IS NOT DISTINCT FROM：

SELECT * FROM numbers JOIN letters 
ON numbers.numbers IS NOT DISTINCT FROM letters.numbers

这也可以与DataFrame API一起使用：

numbersDf.alias("numbers")
  .join(lettersDf.alias("letters"))
  .where("numbers.numbers IS NOT DISTINCT FROM letters.numbers")

（编辑：李大同）

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