如何基于旧的 DataFrame 创建新的 DataFrame？

Question

我有 csv 文件：dbname1.table1.csv:

|target            | source        |source_table                       |relation_type|
 ---------------------------------------------------------------------------------------
avg_ensure_sum_12m | inn_num       | custom_cib_ml_stg.p_overall_part_tend_cust | direct
avg_ensure_sum_12m | protocol_dttm | custom_cib_ml_stg.p_overall_part_tend_cust | direct
avg_ensure_sum_12m | inn_num       | custom_cib_ml_stg.p_overall_part_tend_cust | indirect

此 table 的 csv 格式：

target,source,source_table,relation_type
avg_ensure_sum_12m,inn_num,custom_cib_ml_stg.p_overall_part_tend_cust,direct
avg_ensure_sum_12m,protocol_dttm,custom_cib_ml_stg.p_overall_part_tend_cust,direct
avg_ensure_sum_12m,inn_num,custom_cib_ml_stg.p_overall_part_tend_cust,indirect

然后我通过读取它来创建一个数据框：

 val dfDL = spark.read.option("delimiter", ",")
                     .option("header", true)
                     .csv(file.getPath.toUri.getPath)

现在我需要创建一个基于 dfDL 的新数据框。

新数据框的结构如下所示：

case class DataLink(schema_from: String,
                    table_from: String,
                    column_from: String,
                    link_type: String,
                    schema_to: String,
                    table_to: String,
                    column_to: String)

新DataFrame的字段信息是从一个csv文件中获取的：

pseudocode:
schema_from = source_table.split(".")(0) // Example: custom_cib_ml_stg
table_from  = source_table.split(".")(1) // Example: p_overall_part_tend_cust
column_from = source                     // Example: inn_num
link_type   = relation_type              // Example: direct
schema_to   = "dbname1.table1.csv".split(".")(0) // Example: dbname1
table_to    = "dbname1.table1.csv".split(".")(1) // Example: table1
column_to   = target                             // Example: avg_ensure_sum_12m

我需要创建一个新的数据框。我一个人应付不了。

P.S。我需要这个数据框以便稍后从中创建一个 json 文件。例子 JSON:

[{"schema_from":"custom_cib_ml36_stg",
"table_from":"p_overall_part_tend_cust",
"column_from":"inn_num",
"link_type":"direct",
"schema_to":"dbname1",
"table_to":"table1",
"column_to":"avg_ensure_sum_12m"
},
{"schema_from":"custom_cib_ml36_stg",
"table_from":"p_overall_part_tend_cust",
"column_from":"protocol_dttm",
"link_type":"direct","schema_to":"dbname1",
"table_to":"table1",
"column_to":"avg_ensure_sum_12m"}

我不喜欢我当前的实现：

def readDLFromHDFS(file: LocatedFileStatus): Array[DataLink] = {

    val arrTableName        = file.getPath.getName.split("\.")
    val (schemaTo, tableTo) = (arrTableName(0), arrTableName(1))

    val dfDL = spark.read.option("delimiter", ",")
                         .option("header", true)
                         .csv(file.getPath.toUri.getPath)

    //val sourceTable = dfDL.select("source_table").collect().map(value => value.toString().split("."))

    dfDL.collect.map(row => DataLink(row.getString(2).split("\.")(0),
                                     row.getString(2).split("\.")(1),
                                     row.getString(1),
                                     row.getString(3),
                                     schemaTo,
                                     tableTo,
                                     row.getString(0)))
  }

  def toJSON(dataLinks: Array[DataLink]): Option[JValue] =
    dataLinks.map(Extraction.decompose).reduceOption(_ ++ _)

}

Answer 1

你肯定不想收集，这就违背了在这里使用 spark 的意义。与 Spark 一样，您有很多选择。您可以使用 RDD，但我认为无需在此处切换模式。您只想将自定义逻辑应用于某些列，并最终得到一个仅包含结果列的数据框。

首先，定义一个要应用的UDF：

def convert(target, source, source_table, relation_type) =
  DataLink(source_table.split("\.")(0),
           source_table.split("\.")(1),
           source,
           "dbname1.table1.csv".split(".")(0)
           "dbname1.table1.csv".split(".")(1)
           target))

然后将此函数应用于所有相关列（确保将其包装在 udf 中以使其成为 spark 函数而不是普通的 Scala 函数）并且 select 结果：

df.select(udf(convert)($"target", $"source", $"source_table", $"relation_type"))

如果您想要 DataFrame 包含 7 列的结果：

df.select(
  split(col("source_table"), "\.").getItem(0),
  split(col("source_table"), "\.").getItem(1),
  col("source"),
  lit("dbname1"),
  lit("table1"),
  col("target")
)

您还可以将 .as("column_name") 添加到这 7 列中的每一列。

Answer 2

您可以直接使用数据集。

import spark.implicits._

case class DataLink(schema_from: String,
                    table_from: String,
                    column_from: String,
                    link_type: String,
                    schema_to: String,
                    table_to: String,
                    column_to: String)

val filename = "dbname1.table1.csv"
val df = spark.read.option("header","true").csv("test.csv")
df.show(false)
+------------------+-------------+------------------------------------------+-------------+
|target            |source       |source_table                              |relation_type|
+------------------+-------------+------------------------------------------+-------------+
|avg_ensure_sum_12m|inn_num      |custom_cib_ml_stg.p_overall_part_tend_cust|direct       |
|avg_ensure_sum_12m|protocol_dttm|custom_cib_ml_stg.p_overall_part_tend_cust|direct       |
|avg_ensure_sum_12m|inn_num      |custom_cib_ml_stg.p_overall_part_tend_cust|indirect     |
+------------------+-------------+------------------------------------------+-------------+

df.createOrReplaceTempView("table")

val df2 = spark.sql(s"""
select split(source_table, '[.]')[0] as schema_from
     , split(source_table, '[.]')[1] as table_from
     , source                        as column_from
     , relation_type                 as link_type
     , split('${filename}', '[.]')[0] as schema_to
     , split('${filename}', '[.]')[1] as table_to
     , target                        as column_to
  from table
""").as[DataLink]

df2.show()

+-----------------+--------------------+-------------+---------+---------+--------+------------------+
|      schema_from|          table_from|  column_from|link_type|schema_to|table_to|         column_to|
+-----------------+--------------------+-------------+---------+---------+--------+------------------+
|custom_cib_ml_stg|p_overall_part_te...|      inn_num|   direct|  dbname1|  table1|avg_ensure_sum_12m|
|custom_cib_ml_stg|p_overall_part_te...|protocol_dttm|   direct|  dbname1|  table1|avg_ensure_sum_12m|
|custom_cib_ml_stg|p_overall_part_te...|      inn_num| indirect|  dbname1|  table1|avg_ensure_sum_12m|
+-----------------+--------------------+-------------+---------+---------+--------+------------------+

Answer 3

我的进步... 现在，我可以创建新的 DataFrame，但他只包含 1 列。

val dfDL = spark.read.option("delimiter", ",")
                     .option("header", true)
                     .csv(file.getPath.toUri.getPath)

val convertCase = (target: String, source: String, source_table: String, relation_type: String) =>
                    DataLink(
                      source_table.split("\.")(0),
                      source_table.split("\.")(1),
                      source,
                      relation_type,
                      schemaTo,
                      tableTo,
                      target,
                    )


val udfConvert = udf(convertCase)

val dfForJson  = dfDL.select(udfConvert(col("target"),
                                        col("source"),
                                        col("source_table"),
                                        col("relation_type")))

如何基于旧的 DataFrame 创建新的 DataFrame？

How do I create a new DataFame based on an old DataFrame?

dataframe

scala

apache-spark

apache-spark-sql