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

Asked By: WestCoastProjects

||

Source

Answer 1

Use the method .rdd like this:

rdd = df.rdd

Answered By: dapangmao

Answer 2

@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)

Answered By: kennyut

Answer 3

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))

Answered By: Nilesh

How to convert a DataFrame back to normal RDD in pyspark?

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