scala - 获取 Spark 数据帧中 ArrayType 列的不同元素

Question

我有一个包含 3 列名为 id、feat1 和 feat2 的数据框。 feat1 和 feat2 是字符串数组的形式:

Id, feat1,feat2
------------------
1, ["feat1_1","feat1_2","feat1_3"],[] 

2, ["feat1_2"],["feat2_1","feat2_2"]

3,["feat1_4"],["feat2_3"]

我想获取每个特征列中不同元素的列表，因此输出将是:

distinct_feat1,distinct_feat2
-----------------------------  
["feat1_1","feat1_2","feat1_3","feat1_4"],["feat2_1","feat2_2","feat2_3]

在 Scala 中执行此操作的最佳方法是什么？

Answer 1

您可以使用 collect_set在应用 explode 后找到对应列的不同值函数在每一列上取消嵌套每个单元格中的数组元素。假设您的数据框名为 df :

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

val distinct_df = df.withColumn("feat1", explode(col("feat1"))).
                     withColumn("feat2", explode(col("feat2"))).
                     agg(collect_set("feat1").alias("distinct_feat1"), 
                         collect_set("feat2").alias("distinct_feat2"))

distinct_df.show
+--------------------+--------------------+
|      distinct_feat1|      distinct_feat2|
+--------------------+--------------------+
|[feat1_1, feat1_2...|[, feat2_1, feat2...|
+--------------------+--------------------+


distinct_df.take(1)
res23: Array[org.apache.spark.sql.Row] = Array([WrappedArray(feat1_1, feat1_2, feat1_3, feat1_4),
                                                WrappedArray(, feat2_1, feat2_2, feat2_3)])