Python 创建 Spark 数据帧。无法推断类型的架构:<type 'float'>
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Create Spark DataFrame. Can not infer schema for type: <type 'float'>
提问by Breach
Could someone help me solve this problem I have with Spark DataFrame?
有人可以帮我解决 Spark DataFrame 遇到的这个问题吗?
When I do myFloatRDD.toDF()
I get an error:
当我这样做时,myFloatRDD.toDF()
我收到一个错误:
TypeError: Can not infer schema for type: type 'float'
类型错误:无法推断类型的架构:类型“浮动”
I don't understand why...
我不明白为什么...
Example:
例子:
myFloatRdd = sc.parallelize([1.0,2.0,3.0])
df = myFloatRdd.toDF()
Thanks
谢谢
采纳答案by zero323
SparkSession.createDataFrame
, which is used under the hood, requires an RDD
/ list
of Row
/tuple
/list
/* or dict
pandas.DataFrame
, unless schema with DataType
is provided. Try to convert float to tuple like this:
SparkSession.createDataFrame
,这是发动机罩下使用的,需要一个RDD
/list
的 Row
/ tuple
/ list
/ *或者dict
pandas.DataFrame
,除非用模式DataType
设置。尝试将浮点数转换为元组,如下所示:
myFloatRdd.map(lambda x: (x, )).toDF()
or even better:
甚至更好:
from pyspark.sql import Row
row = Row("val") # Or some other column name
myFloatRdd.map(row).toDF()
To create a DataFrame
from a list of scalars you'll have to use SparkSession.createDataFrame
directly and provide a schema***:
要从DataFrame
标量列表创建一个,您必须SparkSession.createDataFrame
直接使用并提供架构***:
from pyspark.sql.types import FloatType
df = spark.createDataFrame([1.0, 2.0, 3.0], FloatType())
df.show()
## +-----+
## |value|
## +-----+
## | 1.0|
## | 2.0|
## | 3.0|
## +-----+
but for a simple range it would be better to use SparkSession.range
:
但对于一个简单的范围,最好使用SparkSession.range
:
from pyspark.sql.functions import col
spark.range(1, 4).select(col("id").cast("double"))
* No longer supported.
* 不再支持。
** Spark SQL also provides a limited support for schema inference on Python objects exposing __dict__
.
** Spark SQL 还对暴露__dict__
.
*** Supported only in Spark 2.0 or later.
*** 仅在 Spark 2.0 或更高版本中受支持。