使用Pyspark将密集向量转换为数据框 [英] Convert a Dense Vector to a Dataframe using Pyspark

问题描述

首先，我尝试使用下面链接中的所有内容来修复我的错误，但没有一个起作用。

Firstly I tried everything in the link below to fix my error but none of them worked.

如何将密集矢量的RDD转换为pyspark中的DataFrame？

我正在尝试将密集向量与列名一起转换为数据帧（最好是Spark）并遇到问题。

I am trying to convert a dense vector into a dataframe (Spark preferably) along with column names and running into issues.

我的列spark dataframe中的一个向量是使用Vector Assembler创建的向量，现在我想将其转换回一个数据框架，因为我想在向量中的某些变量上创建图。

My column in spark dataframe is a vector that was created using Vector Assembler and I now want to convert it back to a dataframe as I would like to create plots on some of the variables in the vector.

方法1：

from pyspark.ml.linalg import SparseVector, DenseVector
from pyspark.ml.linalg import Vectors

temp=output.select("all_features")
temp.rdd.map(
    lambda row: (DenseVector(row[0].toArray()))
).toDF()

以下是错误

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

方法2：

from pyspark.ml.linalg import VectorUDT
from pyspark.sql.functions import udf
from pyspark.ml.linalg import *

as_ml = udf(lambda v: v.asML() if v is not None else None, VectorUDT())
result = output.withColumn("all_features", as_ml("all_features"))
result.head(5)

错误：

AttributeError: 'numpy.ndarray' object has no attribute 'asML'

I还尝试将数据框转换为Pandas数据框，之后我无法将值拆分为单独的列

I also tried to convert the dataframe into a Pandas dataframe and after that I am not able to split the values into separate columns

方法3：

pandas_df=temp.toPandas()
pandas_df1=pd.DataFrame(pandas_df.all_features.values.tolist())

上面的代码运行良好，但我的数据框中仍然只有一列，所有值都用逗号分隔为列表。

Above code runs fine but I still have only one column in my dataframe with all the values separated by commas as a list.

任何帮助我

编辑：

这是我的临时数据帧的样子。它只有一列all_features。我正在尝试创建一个将所有这些值拆分为单独列的数据框（all_features是使用200列创建的向量）

Here is how my temp dataframe looks like. It just has one column all_features. I am trying to create a dataframe that splits all of these values into separate columns (all_features is a vector that was created using 200 columns)

+--------------------+
|        all_features|
+--------------------+
|[0.01193689934723...|
|[0.04774759738895...|
|[0.0,0.0,0.194417...|
|[0.02387379869447...|
|[1.89796699621085...|
+--------------------+
only showing top 5 rows

预期输出是一个数据帧，其中所有200列都在一个数据帧中分开

Expected output is a dataframe with all 200 columns separated out in a dataframe

+----------------------------+
|        col1| col2| col3|...
+----------------------------+
|0.01193689934723|0.0|0.5049431301173817...
|0.04774759738895|0.0|0.1657316216149636...
|0.0|0.0|7.213126372469...
|0.02387379869447|0.0|0.1866693496827619|...
|1.89796699621085|0.0|0.3192169213385746|...
+----------------------------+
only showing top 5 rows

这是我的熊猫DF输出的样子

Here is how my Pandas DF output looks like

              0
0   [0.011936899347238104, 0.0, 0.5049431301173817...
1   [0.047747597388952415, 0.0, 0.1657316216149636...
2   [0.0, 0.0, 0.19441761495525278, 7.213126372469...
3   [0.023873798694476207, 0.0, 0.1866693496827619...
4   [1.8979669962108585, 0.0, 0.3192169213385746, ...

推荐答案

由于您希望所有功能都位于单独的列中（如我从EDIT中所获得的），因此指向您提供的答案的链接为

Since you want all the features in separate columns (as I got from your EDIT), the link to the answer you provided is not your solution.

尝试一下，

#column_names
temp = temp.rdd.map(lambda x:[float(y) for y in x['all_features']]).toDF(column_names)

编辑：

由于您的温度最初是一个数据帧，您也可以使用此方法而无需将其转换为 rdd ，

Since your temp is originally a dataframe, you can also use this method without converting it to rdd,

import pyspark.sql.functions as F
from pyspark.sql.types import *

splits = [F.udf(lambda val: float(val[i].item()),FloatType()) for i in range(200)]
temp = temp.select(*[s(F.col('all_features')).alias(c) for c,s in zip(column_names,splits)])
temp.show()

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