Question

我想在Apache Spark连接中包含空值。 Spark默认情况下不包含null的行。

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"))

以下是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|
+-------+-------+

Answer 1

Spark提供了一个特殊的NULL安全等式运算符：

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-11111 - Fast null-safe join ）。

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", ), ("456", ), (None, ), ("", )
]).toDF(["numbers"])

letters_df = sc.parallelize([
    ("123", "abc"), ("456", "def"), (None, "zzz"), ("", "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", "456", NA, ""),
  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

这也适用于DataFrame API：

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

Answer 2

val numbers2 = numbersDf.withColumnRenamed("numbers","num1") //rename columns so that we can disambiguate them in the join
val letters2 = lettersDf.withColumnRenamed("numbers","num2")
val joinedDf = numbers2.join(letters2, $"num1" === $"num2" || ($"num1".isNull &&  $"num2".isNull) ,"outer")
joinedDf.select("num1","letters").withColumnRenamed("num1","numbers").show  //rename the columns back to the original names

Answer 3

作为其他答案的补充，对于 PYSPARK <2.3.0 ，您将没有 Column.eqNullSafe ，也没有不与世隔绝的。

只要您为联接查询定义别名，您仍然可以使用sql表达式构建<=>运算符以将其包括在联接中：

from pyspark.sql.types import StringType
import pyspark.sql.functions as F

numbers_df = spark.createDataFrame (["123","456",None,""], StringType()).toDF("numbers")
letters_df = spark.createDataFrame ([("123", "abc"),("456", "def"),(None, "zzz"),("", "hhh") ]).\
    toDF("numbers", "letters")

joined_df = numbers_df.alias("numbers").join(letters_df.alias("letters"),
                                             F.expr('numbers.numbers <=> letters.numbers')).\
    select('letters.*')
joined_df.show()

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

Answer 4

尝试使用以下方法将空行包含在JOIN运算符的结果中：

def nullSafeJoin(leftDF: DataFrame, rightDF: DataFrame, columns: Seq[String], joinType: String): DataFrame = {

    var columnsExpr: Column = leftDF(columns.head) <=> rightDF(columns.head)

    columns.drop(1).foreach(column => {
        columnsExpr = columnsExpr && (leftDF(column) <=> rightDF(column))
    })

    var joinedDF: DataFrame = leftDF.join(rightDF, columnsExpr, joinType)

    columns.foreach(column => {
        joinedDF = joinedDF.drop(leftDF(column))
    })

    joinedDF
}

Answer 5

基于K L的想法，您可以使用foldLeft生成联接列表达式：

def nullSafeJoin(rightDF: DataFrame, columns: Seq[String], joinType: String)(leftDF: DataFrame): DataFrame = 
{

  val colExpr: Column = leftDF(columns.head) <=> rightDF(columns.head)
  val fullExpr = columns.tail.foldLeft(colExpr) { 
    (colExpr, p) => colExpr && leftDF(p) <=> rightDF(p) 
  }

  leftDF.join(rightDF, fullExpr, joinType)
}

然后，您可以像下面这样调用此函数：

aDF.transform(nullSafejoin(bDF, columns, joinType))

Answer 6

基于 timothyzhang 的想法，可以通过删除重复列来进一步改进它：

def dropDuplicateColumns(df: DataFrame, rightDf: DataFrame, cols: Seq[String]): DataFrame 
= cols.foldLeft(df)((df, c) => df.drop(rightDf(c)))

def joinTablesWithSafeNulls(rightDF: DataFrame, leftDF: DataFrame, columns: Seq[String], joinType: String): DataFrame = 
{

val colExpr: Column = leftDF(columns.head) <=> rightDF(columns.head)

val fullExpr = columns.tail.foldLeft(colExpr) {
  (colExpr, p) => colExpr && leftDF(p) <=> rightDF(p)
}

val finalDF = leftDF.join(rightDF, fullExpr, joinType)

val filteredDF = dropDuplicateColumns(finalDF, rightDF, columns)

filteredDF

}

在Apache Spark Join中包含空值

6 个答案: