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Posted to user@spark.apache.org by ca...@free.fr on 2022/02/07 04:09:48 UTC
TypeError: Can not infer schema for type:
>>> rdd = sc.parallelize([3,2,1,4])
>>> rdd.toDF().show()
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
File "/opt/spark/python/pyspark/sql/session.py", line 66, in toDF
return sparkSession.createDataFrame(self, schema, sampleRatio)
File "/opt/spark/python/pyspark/sql/session.py", line 675, in
createDataFrame
return self._create_dataframe(data, schema, samplingRatio,
verifySchema)
File "/opt/spark/python/pyspark/sql/session.py", line 698, in
_create_dataframe
rdd, schema = self._createFromRDD(data.map(prepare), schema,
samplingRatio)
File "/opt/spark/python/pyspark/sql/session.py", line 486, in
_createFromRDD
struct = self._inferSchema(rdd, samplingRatio, names=schema)
File "/opt/spark/python/pyspark/sql/session.py", line 466, in
_inferSchema
schema = _infer_schema(first, names=names)
File "/opt/spark/python/pyspark/sql/types.py", line 1067, in
_infer_schema
raise TypeError("Can not infer schema for type: %s" % type(row))
TypeError: Can not infer schema for type: <class 'int'>
In my pyspark why this fails? I didnt get the way.
Thanks for helps.
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Re: TypeError: Can not infer schema for type:
Posted by Mich Talebzadeh <mi...@gmail.com>.
Absolutely
The reason this error happens is that an rdd is a one dimensional data
structure whilst a data frame has to be 2 dimensional, i.e. we have a
List[Integer] but we need List[Tuple[Integer]].
Try this
>>> rdd = sc.parallelize([3,2,1,4])
>>> df = rdd.map(lambda x: (x,)).toDF()
>>> df.printSchema()
root
|-- _1: long (nullable = true)
>>> from pyspark.sql.functions import col
>>> df.filter((col("_1") > 2)).show()
+---+
| _1|
+---+
| 3|
| 4|
+---+
or create a dataframe with schema defined
>>> from pyspark.sql.functions import col
>>> Schema = StructType([ StructField("ID", IntegerType(), False)])
>>> df = spark.createDataFrame(sc.parallelize([3,2,1,4]).map(lambda x:
(x,)), schema = Schema)
>>> df.filter(col("ID") > 2).show()
+---+
| ID|
+---+
| 3|
| 4|
+---+
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On Mon, 7 Feb 2022 at 04:42, Sean Owen <sr...@gmail.com> wrote:
> You are passing a list of primitives. It expects something like a list of
> tuples, which can each have 1 int if you like.
>
> On Sun, Feb 6, 2022, 10:10 PM <ca...@free.fr> wrote:
>
>> >>> rdd = sc.parallelize([3,2,1,4])
>> >>> rdd.toDF().show()
>> Traceback (most recent call last):
>> File "<stdin>", line 1, in <module>
>> File "/opt/spark/python/pyspark/sql/session.py", line 66, in toDF
>> return sparkSession.createDataFrame(self, schema, sampleRatio)
>> File "/opt/spark/python/pyspark/sql/session.py", line 675, in
>> createDataFrame
>> return self._create_dataframe(data, schema, samplingRatio,
>> verifySchema)
>> File "/opt/spark/python/pyspark/sql/session.py", line 698, in
>> _create_dataframe
>> rdd, schema = self._createFromRDD(data.map(prepare), schema,
>> samplingRatio)
>> File "/opt/spark/python/pyspark/sql/session.py", line 486, in
>> _createFromRDD
>> struct = self._inferSchema(rdd, samplingRatio, names=schema)
>> File "/opt/spark/python/pyspark/sql/session.py", line 466, in
>> _inferSchema
>> schema = _infer_schema(first, names=names)
>> File "/opt/spark/python/pyspark/sql/types.py", line 1067, in
>> _infer_schema
>> raise TypeError("Can not infer schema for type: %s" % type(row))
>> TypeError: Can not infer schema for type: <class 'int'>
>>
>>
>> In my pyspark why this fails? I didnt get the way.
>> Thanks for helps.
>>
>> ---------------------------------------------------------------------
>> To unsubscribe e-mail: user-unsubscribe@spark.apache.org
>>
>>
Re: TypeError: Can not infer schema for type:
Posted by ca...@free.fr.
Thanks for the reply.
It looks strange that in scala shell I can implement this translation:
scala> sc.parallelize(List(3,2,1,4)).toDF.show
+-----+
|value|
+-----+
| 3|
| 2|
| 1|
| 4|
+-----+
But in pyspark i have to write as:
>>> sc.parallelize([3,2,1,4]).map(lambda x:
>>> (x,1)).toDF(['id','count']).show()
+---+-----+
| id|count|
+---+-----+
| 3| 1|
| 2| 1|
| 1| 1|
| 4| 1|
+---+-----+
So there are differences on the implementation of pyspark and scala.
Thanks
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Re: TypeError: Can not infer schema for type:
Posted by Sean Owen <sr...@gmail.com>.
You are passing a list of primitives. It expects something like a list of
tuples, which can each have 1 int if you like.
On Sun, Feb 6, 2022, 10:10 PM <ca...@free.fr> wrote:
> >>> rdd = sc.parallelize([3,2,1,4])
> >>> rdd.toDF().show()
> Traceback (most recent call last):
> File "<stdin>", line 1, in <module>
> File "/opt/spark/python/pyspark/sql/session.py", line 66, in toDF
> return sparkSession.createDataFrame(self, schema, sampleRatio)
> File "/opt/spark/python/pyspark/sql/session.py", line 675, in
> createDataFrame
> return self._create_dataframe(data, schema, samplingRatio,
> verifySchema)
> File "/opt/spark/python/pyspark/sql/session.py", line 698, in
> _create_dataframe
> rdd, schema = self._createFromRDD(data.map(prepare), schema,
> samplingRatio)
> File "/opt/spark/python/pyspark/sql/session.py", line 486, in
> _createFromRDD
> struct = self._inferSchema(rdd, samplingRatio, names=schema)
> File "/opt/spark/python/pyspark/sql/session.py", line 466, in
> _inferSchema
> schema = _infer_schema(first, names=names)
> File "/opt/spark/python/pyspark/sql/types.py", line 1067, in
> _infer_schema
> raise TypeError("Can not infer schema for type: %s" % type(row))
> TypeError: Can not infer schema for type: <class 'int'>
>
>
> In my pyspark why this fails? I didnt get the way.
> Thanks for helps.
>
> ---------------------------------------------------------------------
> To unsubscribe e-mail: user-unsubscribe@spark.apache.org
>
>