How to convert a DataFrame back to normal RDD in pyspark

Question

I need to use the    rdd  partitionBy npartitions  custom partitioner    method that is not available on the DataFrame   All of the DataFrame methods refer only to DataFrame results   So then how to create an RDD from the DataFrame data   Note  this is a change  in 1 3 0  from 1 2 0   Update from the answer from  dpangmao  the method is  rdd  I was interested to understand if  a  it were public and  b  what are the performance implications   Well  a  is yes and  b  - well you can see here that there are significant perf implications  a new RDD must be created  by invoking mapPartitions    In dataframe py  note the file name changed as well  was sql py     property def rdd self               Return the content of the  class  DataFrame  as an  class  RDD      of  class  Row  s              if not hasattr self    lazy rdd            jrdd   self  jdf javaToPython           rdd   RDD jrdd  self sql ctx  sc  BatchedSerializer PickleSerializer             schema   self schema          def applySchema it               cls    create cls schema              return itertools imap cls  it           self  lazy rdd   rdd mapPartitions applySchema       return self  lazy rdd

User · Accepted Answer

dapangmao s answer works  but it doesn t give the regular spark RDD  it returns a Row object  If you want to have the regular RDD format   Try this   rdd   df rdd map tuple    or  rdd   df rdd map list

User · Answer

Use the method  rdd like this   rdd   df rdd

User · Answer

Answer given by kennyut Kistian works very well but to get exact RDD like output when RDD consist of list of attributes e g   1 2 3 4  we can use flatmap command as below   rdd   df rdd flatMap list  or  rdd   df rdd flatmap lambda x  list x

[python] How to convert a DataFrame back to normal RDD in pyspark?

Examples related to python

Examples related to apache-spark

Examples related to pyspark