比较Scala Spark中的两个数组列
我有一个下面给出的数据帧格式比较Scala Spark中的两个数组列,scala,apache-spark,array-column,Scala,Apache Spark,Array Column,我有一个下面给出的数据帧格式 movieId1 | genreList1 | genreList2 -------------------------------------------------- 1 |[Adventure,Comedy] |[Adventure] 2 |[Animation,Drama,War] |[War,Drama] 3 |[Adventure,Drama] |[Dra
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")))
但这会抛出org.apache.spark.SparkException:无法执行用户定义的函数。
错误。关于如何解决这个问题的任何想法。
注:上述函数(
intersect\u check
)适用于数组
S。我们可以定义一个
udf
,它计算两个数组
列之间的交集
的长度,并检查它是否等于第二列的长度。如果是,则第二个数组是第一个数组的子集
另外,udf
的输入需要是classWrappedArray[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")
我们可以定义一个
udf
,它计算两个数组
列之间的交集
的长度,并检查它是否等于第二列的长度。如果是,则第二个数组是第一个数组的子集
另外,udf
的输入需要是classWrappedArray[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")
下面是使用
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()
希望这有帮助 以下是使用
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()
希望这有帮助 一种解决方案可能是利用spark array内置函数:
genreList2
是genreList1
的子集,如果两者的交集等于genreList2
。在下面的代码中,添加了一个sort_array
操作,以避免两个排序不同但元素相同的数组之间出现不匹配
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($"genreList2")
)
.cast("integer")
)
.show()
输出是
+--------+--------------------+------------+----+
|movieId1| genreList1| genreList2|flag|
+--------+--------------------+------------+----+
| 1| [Adventure, Comedy]| [Adventure]| 1|
| 2|[Animation, Drama...|[War, Drama]| 1|
| 3| [Adventure, Drama]|[Drama, War]| 0|
+--------+--------------------+------------+----+
一种解决方案可能是利用spark阵列内置函数:
genreList2
是genreList1
的子集,如果两者的交集等于genreList2
。在下面的代码中,添加了一个sort_array
操作,以避免两个排序不同但元素相同的数组之间出现不匹配
val spark = {
SparkSession
.builder()
.master("local")
.appName("test")
.getOrCreate()
}
import spark.implicits._
import org.apache.spark.sql._
import org.apache.spark.sql.functions._
val df = Seq(
(1, Array("Adventure","Comedy"), Array("Adventure")),
(2, Array("Animation","Drama","War"), Array("War","Drama")),
(3, Array("Adventure","Drama"), Array("Drama","War"))
).toDF("movieId1", "genreList1", "genreList2")
df
.withColumn("flag",
sort_array(array_intersect($"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|
+--------+--------------------+------------+----+
请检查同一数据集的两个示例,它们不匹配。谢谢..更新..请检查同一数据集的两个示例,它们不匹配。谢谢..更新。。