如何拆分输入文件名并在 spark 数据框列中添加特定值

Question

这就是我在 spark 数据框中加载 csv 文件的方式

val sqlContext = new org.apache.spark.sql.SQLContext(sc)
import sqlContext.implicits._

import org.apache.spark.{ SparkConf, SparkContext }
import java.sql.{Date, Timestamp}
import org.apache.spark.sql.Row
import org.apache.spark.sql.types._
import org.apache.spark.sql.functions.udf



val get_cus_val = spark.udf.register("get_cus_val", (filePath: String) => filePath.split("\.")(4))

val df1With_ = df.toDF(df.columns.map(_.replace(".", "_")): _*)
val column_to_keep = df1With_.columns.filter(v => (!v.contains("^") && !v.contains("!") && !v.contains("_c"))).toSeq
val df1result = df1With_.select(column_to_keep.head, column_to_keep.tail: _*)
val df1Final=df1result.withColumn("DataPartition", lit(null: String))

这是我的输入文件名之一的示例。

Fundamental.FinancialLineItem.FinancialLineItem.SelfSourcedPrivate.CUS.1.2017-09-07-1056.Full

Fundamental.FinancialLineItem.FinancialLineItem.Japan.CUS.1.2017-09-07-1056.Full.txt

现在我想读取这个文件并用“.”分割它。运算符，然后将 CUS 添加为新列以代替 DataPartition 。

我可以不使用任何 UDF 吗？

这是现有数据框的模式

root
 |-- LineItem_organizationId: long (nullable = true)
 |-- LineItem_lineItemId: integer (nullable = true)
 |-- StatementTypeCode: string (nullable = true)
 |-- LineItemName: string (nullable = true)
 |-- LocalLanguageLabel: string (nullable = true)
 |-- FinancialConceptLocal: string (nullable = true)
 |-- FinancialConceptGlobal: string (nullable = true)
 |-- IsDimensional: boolean (nullable = true)
 |-- InstrumentId: string (nullable = true)
 |-- LineItemSequence: string (nullable = true)
 |-- PhysicalMeasureId: string (nullable = true)
 |-- FinancialConceptCodeGlobalSecondary: string (nullable = true)
 |-- IsRangeAllowed: boolean (nullable = true)
 |-- IsSegmentedByOrigin: boolean (nullable = true)
 |-- SegmentGroupDescription: string (nullable = true)
 |-- SegmentChildDescription: string (nullable = true)
 |-- SegmentChildLocalLanguageLabel: string (nullable = true)
 |-- LocalLanguageLabel_languageId: integer (nullable = true)
 |-- LineItemName_languageId: integer (nullable = true)
 |-- SegmentChildDescription_languageId: integer (nullable = true)
 |-- SegmentChildLocalLanguageLabel_languageId: integer (nullable = true)
 |-- SegmentGroupDescription_languageId: integer (nullable = true)
 |-- SegmentMultipleFundbDescription: string (nullable = true)
 |-- SegmentMultipleFundbDescription_languageId: integer (nullable = true)
 |-- IsCredit: boolean (nullable = true)
 |-- FinancialConceptLocalId: integer (nullable = true)
 |-- FinancialConceptGlobalId: integer (nullable = true)
 |-- FinancialConceptCodeGlobalSecondaryId: string (nullable = true)
 |-- FFAction: string (nullable = true)

根据建议的答案更新代码

    val sqlContext = new org.apache.spark.sql.SQLContext(sc)
    import sqlContext.implicits._

    import org.apache.spark.{ SparkConf, SparkContext }
    import java.sql.{Date, Timestamp}
    import org.apache.spark.sql.Row
    import org.apache.spark.sql.types._
    import org.apache.spark.sql.functions.udf
    import org.apache.spark.sql.functions.{input_file_name, regexp_extract}

spark.udf.register("get_cus_val", (filePath: String) => filePath.split("\.")(4))

import org.apache.spark.sql.functions.input_file_name

val df = sqlContext.read.format("csv").option("header", "true").option("delimiter", "|").option("inferSchema","true").load("s3://trfsdisu/SPARK/FinancialLineItem/MAIN")

val df1With_ = df.toDF(df.columns.map(_.replace(".", "_")): _*)
val column_to_keep = df1With_.columns.filter(v => (!v.contains("^") && !v.contains("!") && !v.contains("_c"))).toSeq
val df1result = df1With_.select(column_to_keep.head, column_to_keep.tail: _*)

df1result.withColumn("cus_val", get_cus_val(input_file_name))

df1result.printSchema()

Answer 1

您可以使用预定义的 UDF 获取文件名，即 input_file_name()，之后您可以创建一个 UDF 来提取 CUS 或使用 regexp_extract wo自定义函数。

使用 regexp_extract wo UDF regex usage here

import org.apache.spark.sql.functions.input_file_name
import org.apache.spark.sql.functions.regexp_extract

df.withColumn("cus_val", 
  regexp_extract(input_file_name, "\.(\w+)\.[0-9]+\.", 1))

使用自定义 UDF

import org.apache.spark.sql.functions.udf

val get_cus_val = udf(filePath: String => filePath.split("\.")(4))

import org.apache.spark.sql.functions.input_file_name

df.withColumn("cus_val", get_cus_val(input_file_name))

如何拆分输入文件名并在 spark 数据框列中添加特定值

How to split input file name and add specific value in the spark data frame column

scala

apache-spark

spark-dataframe

spark-csv