ueshin commented on code in PR #45699:
URL: https://github.com/apache/spark/pull/45699#discussion_r1552202465


##########
python/pyspark/sql/connect/session.py:
##########
@@ -418,6 +425,28 @@ def createDataFrame(
             # If no schema supplied by user then get the names of columns only
             if schema is None:
                 _cols = [str(x) if not isinstance(x, str) else x for x in 
data.columns]
+                infer_pandas_dict_as_map = (
+                    
str(self.conf.get("spark.sql.execution.pandas.inferPandasDictAsMap")).lower()
+                    == "true"
+                )
+                if infer_pandas_dict_as_map:
+                    struct = StructType()
+                    pa_schema = pa.Schema.from_pandas(data)
+                    spark_type: Union[MapType, DataType]
+                    for field in pa_schema:
+                        field_type = field.type
+                        if isinstance(field_type, pa.StructType):
+                            if len(field_type) == 0:
+                                raise PySparkValueError(
+                                    error_class="CANNOT_INFER_EMPTY_SCHEMA",
+                                    message_parameters={},
+                                )
+                            arrow_type = field_type.field(0).type

Review Comment:
   The API `StructType.field` seems to be available since pyarrow `10.0`.
   https://arrow.apache.org/docs/10.0/python/generated/pyarrow.StructType.html
   
   We may want to bump up the minimum pyarrow version, or we should avoid this 
API?



-- 
This is an automated message from the Apache Git Service.
To respond to the message, please log on to GitHub and use the
URL above to go to the specific comment.

To unsubscribe, e-mail: [email protected]

For queries about this service, please contact Infrastructure at:
[email protected]


---------------------------------------------------------------------
To unsubscribe, e-mail: [email protected]
For additional commands, e-mail: [email protected]

Reply via email to