2

我有一个格式如下的数据框。

movieId1 | genreList1              | genreList2
--------------------------------------------------
1        |[Adventure,Comedy]       |[Adventure]
2        |[Animation,Drama,War]    |[War,Drama]
3        |[Adventure,Drama]        |[Drama,War]

并尝试创建另一个标志列，显示genreList2 是否是genreList1 的子集

movieId1 | genreList1              | genreList2        | Flag
---------------------------------------------------------------
1        |[Adventure,Comedy]       | [Adventure]       |1
2        |[Animation,Drama,War]    | [War,Drama]       |1
3        |[Adventure,Drama]        | [Drama,War]       |0

我试过这个

def intersect_check(a: Array[String], b: Array[String]): Int = {
  if (b.sameElements(a.intersect(b))) { return 1 } 
  else { return 2 }
}

def intersect_check_udf =
  udf((colvalue1: Array[String], colvalue2: Array[String]) => intersect_check(colvalue1, colvalue2))

data = data.withColumn("Flag", intersect_check_udf(col("genreList1"), col("genreList2")))

但这会引发org.apache.spark.SparkException: Failed to execute user defined function.错误。关于如何解决这个问题的任何想法。PS：上述函数 ( intersect_check) 适用于Arrays。

4

3 回答 3

5

我们可以定义一个udf计算intersection两Array列之间的长度并检查它是否等于第二列的长度。如果是这样，则第二个数组是第一个数组的子集。

此外，您的输入udf需要是 class WrappedArray[String]，而不是Array[String]：

import scala.collection.mutable.WrappedArray
import org.apache.spark.sql.functions.col

val same_elements = udf { (a: WrappedArray[String], 
                           b: WrappedArray[String]) => 
  if (a.intersect(b).length == b.length){ 1 }else{ 0 }  
}

df.withColumn("test",same_elements(col("genreList1"),col("genreList2")))
  .show(truncate = false)
+--------+-----------------------+------------+----+
|movieId1|genreList1             |genreList2  |test|
+--------+-----------------------+------------+----+
|1       |[Adventure, Comedy]    |[Adventure] |1   |
|2       |[Animation, Drama, War]|[War, Drama]|1   |
|3       |[Adventure, Drama]     |[Drama, War]|0   |
+--------+-----------------------+------------+----+

数据

val df = List((1,Array("Adventure","Comedy"), Array("Adventure")),
              (2,Array("Animation","Drama","War"), Array("War","Drama")),
              (3,Array("Adventure","Drama"),Array("Drama","War"))).toDF("movieId1","genreList1","genreList2")
于 2017-05-24T12:53:56.570 回答
3

这是使用转换的解决方案subsetOf

  val spark =
    SparkSession.builder().master("local").appName("test").getOrCreate()

  import spark.implicits._

  val data = spark.sparkContext.parallelize(
  Seq(
    (1,Array("Adventure","Comedy"),Array("Adventure")),
  (2,Array("Animation","Drama","War"),Array("War","Drama")),
  (3,Array("Adventure","Drama"),Array("Drama","War"))
  )).toDF("movieId1", "genreList1", "genreList2")


  val subsetOf = udf((col1: Seq[String], col2: Seq[String]) => {
    if (col2.toSet.subsetOf(col1.toSet)) 1 else 0
  })

  data.withColumn("flag", subsetOf(data("genreList1"), data("genreList2"))).show()

希望这可以帮助！

于 2017-05-24T13:16:29.820 回答
0

一种解决方案可能是利用 spark 数组内置函数：genreList2如果genreList1两者之间的交集等于genreList2. 在下面的代码中，sort_array添加了一个操作以避免两个具有不同排序但相同元素的数组之间的不匹配。

val spark = {
    SparkSession
    .builder()
    .master("local")
    .appName("test")
    .getOrCreate()
}

import spark.implicits._
import org.apache.spark.sql._
import org.apache.spark.sql.functions._

val df = Seq(
    (1, Array("Adventure","Comedy"), Array("Adventure")),
    (2, Array("Animation","Drama","War"), Array("War","Drama")),
    (3, Array("Adventure","Drama"), Array("Drama","War"))
).toDF("movieId1", "genreList1", "genreList2")

df
.withColumn("flag",
 sort_array(array_intersect($"genreList1",$"genreList2"))
 .equalTo(
   sort_array($"genreList2")
 )
.cast("integer")
)
.show()

输出是

+--------+--------------------+------------+----+
|movieId1|          genreList1|  genreList2|flag|
+--------+--------------------+------------+----+
|       1| [Adventure, Comedy]| [Adventure]|   1|
|       2|[Animation, Drama...|[War, Drama]|   1|
|       3|  [Adventure, Drama]|[Drama, War]|   0|
+--------+--------------------+------------+----+
于 2021-04-09T11:31:23.770 回答