pyspark:ValueError:推断后无法确定某些类型

Eda*_*ame 17 python python-2.7 pandas pyspark spark-dataframe

我有一个熊猫数据框my_df,并my_df.dtypes给我们:

ts              int64
fieldA         object
fieldB         object
fieldC         object
fieldD         object
fieldE         object
dtype: object
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然后我尝试通过以下操作将pandas数据帧my_df转换为spark数据框:

spark_my_df = sc.createDataFrame(my_df)
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但是,我收到以下错误:

ValueErrorTraceback (most recent call last)
<ipython-input-29-d4c9bb41bb1e> in <module>()
----> 1 spark_my_df = sc.createDataFrame(my_df)
      2 spark_my_df.take(20)

/usr/local/spark-latest/python/pyspark/sql/session.py in createDataFrame(self, data, schema, samplingRatio)
    520             rdd, schema = self._createFromRDD(data.map(prepare), schema, samplingRatio)
    521         else:
--> 522             rdd, schema = self._createFromLocal(map(prepare, data), schema)
    523         jrdd = self._jvm.SerDeUtil.toJavaArray(rdd._to_java_object_rdd())
    524         jdf = self._jsparkSession.applySchemaToPythonRDD(jrdd.rdd(), schema.json())

/usr/local/spark-latest/python/pyspark/sql/session.py in _createFromLocal(self, data, schema)
    384 
    385         if schema is None or isinstance(schema, (list, tuple)):
--> 386             struct = self._inferSchemaFromList(data)
    387             if isinstance(schema, (list, tuple)):
    388                 for i, name in enumerate(schema):

/usr/local/spark-latest/python/pyspark/sql/session.py in _inferSchemaFromList(self, data)
    318         schema = reduce(_merge_type, map(_infer_schema, data))
    319         if _has_nulltype(schema):
--> 320             raise ValueError("Some of types cannot be determined after inferring")
    321         return schema
    322 

ValueError: Some of types cannot be determined after inferring
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有谁知道上述错误是什么意思?谢谢!

Gre*_*ogy 20

为了推断字段类型,PySpark查看每个字段中的非none记录.如果某个字段只有None记录,则PySpark无法推断该类型并会引发该错误.

手动定义架构将解决该问题

>>> from pyspark.sql.types import StructType, StructField, StringType
>>> schema = StructType([StructField("foo", StringType(), True)])
>>> df = spark.createDataFrame([[None]], schema=schema)
>>> df.show()
+----+
|foo |
+----+
|null|
+----+
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  • 我可以只给出整个 None 列的架构并跳过其余列吗? (3认同)

Aka*_*all 11

为了解决这个问题,您可以提供自己定义的架构。

例如:

要重现错误:

>>> df = spark.createDataFrame([[None, None]], ["name", "score"])
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要修复错误:

>>> from pyspark.sql.types import StructType, StructField, StringType, DoubleType
>>> schema = StructType([StructField("name", StringType(), True), StructField("score", DoubleType(), True)])
>>> df = spark.createDataFrame([[None, None]], schema=schema)
>>> df.show()
+----+-----+
|name|score|
+----+-----+
|null| null|
+----+-----+
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  • 如果我们有超过 2 列,并且只有 1 列完全为空,是否有更好的优雅方法来传递架构,而无需为所有列显式定义架构? (3认同)

Aar*_*son 8

我遇到了同样的问题,如果您不需要空列,您可以简单地将它们从 pandas 数据框中删除,然后再导入到 Spark:

my_df = my_df.dropna(axis='columns', how='all') # Drops columns with all NA values
spark_my_df = sc.createDataFrame(my_df)
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rju*_*ney 7

如果您使用的是RDD[Row].toDF()monkey-patched 方法,您可以在推断类型时增加样本比率以检查超过 100 条记录:

# Set sampleRatio smaller as the data size increases
my_df = my_rdd.toDF(sampleRatio=0.01)
my_df.show()
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假设您的 RDD 中的所有字段中都有非空行,那么当您将其增加到sampleRatio1.0时,将更有可能找到它们。