How to rename my JSON generated by pyspark?

后端 未结 1 1149
花落未央
花落未央 2021-01-07 13:47

When i write my JSON file with

dataframe.coalesce(1).write.format(\'json\')

on pyspark im not able to change the name of file in the partit

1条回答
  •  醉梦人生
    2021-01-07 13:57

    In spark we can't control name of the file written to the directory.

    First write the data to the HDFS directory then For changing the name of file we need to use HDFS api.

    Example:

    In Pyspark:

    l=[("a",1)]
    ll=["id","sa"]
    df=spark.createDataFrame(l,ll)
    
    hdfs_dir = "/folder/" #your hdfs directory
    new_filename="my_name.json" #new filename
    
    df.coalesce(1).write.format("json").mode("overwrite").save(hdfs_dir)
    
    fs = spark._jvm.org.apache.hadoop.fs.FileSystem.get(spark._jsc.hadoopConfiguration())
    
    #list files in the directory
    
    list_status = fs.listStatus(spark._jvm.org.apache.hadoop.fs.Path(hdfs_dir))
    
    #filter name of the file starts with part-
    
    file_name = [file.getPath().getName() for file in list_status if file.getPath().getName().startswith('part-')][0]
    
    #rename the file
    
    fs.rename(Path(hdfs_dir+''+file_name),Path(hdfs_dir+''+new_filename))
    

    In case if you want to delete success files in the directory use fs.delete to delete _Success files.

    In Scala:

    val df=Seq(("a",1)).toDF("id","sa")
    df.show(false)
    
    import org.apache.hadoop.fs._
    
    val hdfs_dir = "/folder/"
    val new_filename="new_json.json"
    
    df.coalesce(1).write.mode("overwrite").format("json").save(hdfs_dir)
    
    val fs=FileSystem.get(sc.hadoopConfiguration)
    val f=fs.globStatus(new Path(s"${hdfs_dir}" + "*")).filter(x => x.getPath.getName.toString.startsWith("part-")).map(x => x.getPath.getName).mkString
    
    fs.rename(new Path(s"${hdfs_dir}${f}"),new Path(s"${hdfs_dir}${new_filename}"))
    
    fs.delete(new Path(s"${hdfs_dir}" + "_SUCCESS"))
    

    0 讨论(0)
提交回复
热议问题