需要获取同一文件的scala中不同ID的总数

Question

val someData = Seq(
  Row("test.zip", "abc1","A"),
  Row("test.zip", "abc1","B"),
  Row("test.zip", "abc3","C")  
  )

val someSchema = List(
  StructField("file_name", StringType, true),
  StructField("id", StringType, true),
  StructField("chart_char", StringType, true) 
)

val someDF = spark.createDataFrame(
  spark.sparkContext.parallelize(someData),
  StructType(someSchema)
)

需要以下结果，即相应文件名的不同ID的总数

预期结果：

file_name,count test.zip 2

Answer 1

val df = Seq(
  ("test.zip", "abc1","A"),
  ("test.zip", "abc1","B"),
  ("test.zip", "abc3","C"),
  ("test2.zip", "abc1","C")

).toDF("file_name","id","chart_char")

df.show()

//+---------+----+----------+
//|file_name|  id|chart_char|
//+---------+----+----------+
//| test.zip|abc1|         A|
//| test.zip|abc1|         B|
//| test.zip|abc3|         C|
//|test2.zip|abc1|         C|
//+---------+----+----------+

df.select("file_name", "id")
  .distinct()
  .groupBy("file_name")
  .count
  .show

//+---------+-----+
//|file_name|count|
//+---------+-----+
//| test.zip|    2|
//|test2.zip|    1|
//+---------+-----+