从pyspark.sql中的列表创建数据框 [英] Create a dataframe from a list in pyspark.sql

问题描述

我在联系中完全迷失了.现在我有一个列表li

I am totally lost in a wired situation. Now I have a list li

li = example_data.map(lambda x: get_labeled_prediction(w,x)).collect()
print li, type(li)

输出就像

[(0.0, 59.0), (0.0, 51.0), (0.0, 81.0), (0.0, 8.0), (0.0, 86.0), (0.0, 86.0), (0.0, 60.0), (0.0, 54.0), (0.0, 54.0), (0.0, 84.0)] <type 'list'>

当我尝试从该列表创建数据框

When I try to create a dataframe from this list

m = sqlContext.createDataFrame(l, ["prediction", "label"])

它抛出了错误消息

TypeError                                 Traceback (most recent call last)
<ipython-input-90-4a49f7f67700> in <module>()
 56 l = example_data.map(lambda x: get_labeled_prediction(w,x)).collect()
 57 print l, type(l)
---> 58 m = sqlContext.createDataFrame(l, ["prediction", "label"])
 59 '''
 60 g = example_data.map(lambda x:gradient_summand(w, x)).sum()

/databricks/spark/python/pyspark/sql/context.py in createDataFrame(self, data, schema, samplingRatio)
423             rdd, schema = self._createFromRDD(data, schema, samplingRatio)
424         else:
--> 425             rdd, schema = self._createFromLocal(data, schema)
426         jrdd = self._jvm.SerDeUtil.toJavaArray(rdd._to_java_object_rdd())
427         jdf = self._ssql_ctx.applySchemaToPythonRDD(jrdd.rdd(), schema.json())

/databricks/spark/python/pyspark/sql/context.py in _createFromLocal(self, data, schema)
339 
340         if schema is None or isinstance(schema, (list, tuple)):
--> 341             struct = self._inferSchemaFromList(data)
342             if isinstance(schema, (list, tuple)):
343                 for i, name in enumerate(schema):

/databricks/spark/python/pyspark/sql/context.py in _inferSchemaFromList(self, data)
239             warnings.warn("inferring schema from dict is deprecated,"
240                           "please use pyspark.sql.Row instead")
--> 241         schema = reduce(_merge_type, map(_infer_schema, data))
242         if _has_nulltype(schema):
243             raise ValueError("Some of types cannot be determined after inferring")

/databricks/spark/python/pyspark/sql/types.py in _infer_schema(row)
831         raise TypeError("Can not infer schema for type: %s" % type(row))
832 
--> 833     fields = [StructField(k, _infer_type(v), True) for k, v in items]
834     return StructType(fields)
835 

/databricks/spark/python/pyspark/sql/types.py in _infer_type(obj)
808             return _infer_schema(obj)
809         except TypeError:
--> 810             raise TypeError("not supported type: %s" % type(obj))
811 
812 

TypeError: not supported type: <type 'numpy.float64'>

但是当我在行中硬编码此列表时

But when I hard code this list in line

tt = sqlContext.createDataFrame([(0.0, 59.0), (0.0, 51.0), (0.0, 81.0), (0.0, 8.0), (0.0, 86.0), (0.0, 86.0), (0.0, 60.0), (0.0, 54.0), (0.0, 54.0), (0.0, 84.0)], ["prediction", "label"])
tt.collect()

效果很好.

[Row(prediction=0.0, label=59.0),
 Row(prediction=0.0, label=51.0),
 Row(prediction=0.0, label=81.0),
 Row(prediction=0.0, label=8.0),
 Row(prediction=0.0, label=86.0),
 Row(prediction=0.0, label=86.0),
 Row(prediction=0.0, label=60.0),
 Row(prediction=0.0, label=54.0),
 Row(prediction=0.0, label=54.0),
 Row(prediction=0.0, label=84.0)]

什么原因导致了此问题以及如何解决?任何提示将不胜感激.

what caused this problem and how to fix it? Any hint will be appreciated.

推荐答案

您有一个list of float64，我认为它不喜欢这种类型.另一方面，当您进行硬编码时，它只是一个list of float.
这是一个问题，其中包含有关如何转换的答案从numpy的数据类型到python的本地数据类型.

You have a list of float64 and I think it doesn't like that type. On the other hand, when you hard code it it's just a list of float.
Here is a question with an answer that goes over on how to convert from numpy's datatype to python's native ones.

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