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 3 Answers
Use the method .rdd like this:
rdd = df.rdd @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) 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)) 
