cloud-fan commented on a change in pull request #23534: [SPARK-26610][PYTHON]
Fix inconsistency between toJSON Method in Python and Scala.
URL: https://github.com/apache/spark/pull/23534#discussion_r247466299
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File path: python/pyspark/sql/dataframe.py
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@@ -109,15 +109,18 @@ def stat(self):
@ignore_unicode_prefix
@since(1.3)
def toJSON(self, use_unicode=True):
- """Converts a :class:`DataFrame` into a :class:`RDD` of string.
+ """Converts a :class:`DataFrame` into a :class:`DataFrame` of JSON
string.
- Each row is turned into a JSON document as one element in the returned
RDD.
+ Each row is turned into a JSON document as one element in the returned
DataFrame.
>>> df.toJSON().first()
- u'{"age":2,"name":"Alice"}'
+ Row(value=u'{"age":2,"name":"Alice"}')
"""
- rdd = self._jdf.toJSON()
- return RDD(rdd.toJavaRDD(), self._sc, UTF8Deserializer(use_unicode))
+ jdf = self._jdf.toJSON()
+ if self.sql_ctx._conf.pysparkDataFrameToJSONShouldReturnDataFrame():
+ return DataFrame(jdf, self.sql_ctx)
+ else:
+ return RDD(jdf.toJavaRDD(), self._sc,
UTF8Deserializer(use_unicode))
Review comment:
So which way is more consistent?
If we return DataFrame here, scala side can do `df.toJSON().map(value =>
value.xxx)` and python side needs to do `df.toJSON().map(lambda row:
row.value.xxx)`.
If we return RDD here, scala side can do `df.toJSON().select(...)` and
python needs to do `df.toJSON().toDF().select(...)`
I agree that there is no perfect way since pyspark has no dataset, we should
pick a better but imperfect way.
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