UDF 仅从 Spark SQL 中的路径中提取文件名

UDF to extract only the file name from path in Spark SQL

Apache Spark 中有一个 input_file_name 函数,我用它来将新列添加到数据集中,其中包含当前正在处理的文件的名称。

问题是我想以某种方式将此函数自定义为 return 仅文件名,省略 s3 上的完整路径。

目前,我正在使用 map 函数替换第二步中的路径:

val initialDs = spark.sqlContext.read
.option("dateFormat", conf.dateFormat)
.schema(conf.schema)
.csv(conf.path).withColumn("input_file_name", input_file_name)
...
...
def fromFile(fileName: String): String = {
  val baseName: String = FilenameUtils.getBaseName(fileName)
  val tmpFileName: String = baseName.substring(0, baseName.length - 8) //here is magic conversion ;)
  this.valueOf(tmpFileName)
}

但我想使用类似

的东西
val initialDs = spark.sqlContext.read
    .option("dateFormat", conf.dateFormat)
    .schema(conf.schema)
    .csv(conf.path).withColumn("input_file_name", **customized_input_file_name_function**)

在 Scala 中:

#register udf
spark.udf
  .register("get_only_file_name", (fullPath: String) => fullPath.split("/").last)

#use the udf to get last token(filename) in full path
val initialDs = spark.read
  .option("dateFormat", conf.dateFormat)
  .schema(conf.schema)
  .csv(conf.path)
  .withColumn("input_file_name", get_only_file_name(input_file_name))

编辑:在Java中根据评论

#register udf
spark.udf()
  .register("get_only_file_name", (String fullPath) -> {
     int lastIndex = fullPath.lastIndexOf("/");
     return fullPath.substring(lastIndex, fullPath.length - 1);
    }, DataTypes.StringType);

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

#use the udf to get last token(filename) in full path
Dataset<Row> initialDs = spark.read()
  .option("dateFormat", conf.dateFormat)
  .schema(conf.schema)
  .csv(conf.path)
  .withColumn("input_file_name", get_only_file_name(input_file_name()));

借用一个相关问题,下面的方法更便携,不需要自定义UDF。

Spark SQL 代码段: reverse(split(path, '/'))[0]

Spark SQL 示例:

WITH sample_data as (
SELECT 'path/to/my/filename.txt' AS full_path
)
SELECT
      full_path
    , reverse(split(full_path, '/'))[0] as basename
FROM sample_data

解释: split() 函数将路径分成块,reverse() 将最后一项(文件名)放在数组前面,以便 [0] 可以只提取文件名。

这里有完整的代码示例:

  spark.sql(
    """
      |WITH sample_data as (
      |    SELECT 'path/to/my/filename.txt' AS full_path
      |  )
      |  SELECT
      |  full_path
      |  , reverse(split(full_path, '/'))[0] as basename
      |  FROM sample_data
      |""".stripMargin).show(false)

结果:

+-----------------------+------------+
|full_path              |basename    |
+-----------------------+------------+
|path/to/my/filename.txt|filename.txt|
+-----------------------+------------+

commons io is natural/easiest import in spark means(不需要添加额外的依赖...)

import org.apache.commons.io.FilenameUtils
getBaseName(String fileName)

从完整文件名中获取基本名称,减去完整路径和扩展名。

val baseNameOfFile = udf((longFilePath: String) => FilenameUtils.getBaseName(longFilePath))

用法就像...

yourdataframe.withColumn("shortpath" ,baseNameOfFile(yourdataframe("input_file_name")))
.show(1000,false)