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

python apache-spark pyspark

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

gsamaras

Use the method .rdd like this:

rdd = df.rdd

yes you are correct. I updated the OP after digging deeper into this.

yes but it convert to org.apache.spark.rdd.RDD[org.apache.spark.sql.Row] but not org.apache.spark.rdd.RDD[string]

Technically, it's a property: spark.apache.org/docs/latest/api/python/…

Kristian

@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)

rdd = df.rdd.map(list)

This should be the default behaviour imo when calling df.rdd

This is probably a more precise answer actually

What is the df, how to initialize it?

@DavidWei some Dataframe instance, so whatever variable your dataframe is assigned to

What are tuple and list ?

Nilesh

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))

This looks like a helpful contribution.

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

Follow WeChat

Want to stay one step ahead of the latest teleworks?

相似问题

Platform

Support

Contact US