如何从 org.apache.spark.mllib.linalg.VectorUDT 转换为 ml.linalg.VectorUDT [英] How to convert from org.apache.spark.mllib.linalg.VectorUDT to ml.linalg.VectorUDT

问题描述

我使用的是 Spark cluster 2.0，我想将向量从 org.apache.spark.mllib.linalg.VectorUDT 转换为 org.apache.spark.ml.linalg.VectorUDT.

I am using Spark cluster 2.0 and I would like to convert a vector from org.apache.spark.mllib.linalg.VectorUDT to org.apache.spark.ml.linalg.VectorUDT.

# Import LinearRegression class
from pyspark.ml.regression import LinearRegression

# Define LinearRegression algorithm
lr = LinearRegression()

modelA = lr.fit(data, {lr.regParam:0.0})

错误:

你的要求失败:列特征的类型必须是 org.apache.spark.ml.linalg.VectorUDT@3bfc3ba7，但实际上是 org.apache.spark.mllib.linalg.VectorUDT@f71b0bce.'

u'requirement failed: Column features must be of type org.apache.spark.ml.linalg.VectorUDT@3bfc3ba7 but was actually org.apache.spark.mllib.linalg.VectorUDT@f71b0bce.'

任何想法我将如何在向量类型之间进行这种转换.

Any thoughts how would I do this conversion between vector types.

非常感谢.

推荐答案

在 PySpark 中，您需要一个或 map over RDD.让我们使用第一个选项.首先是几个导入:

In PySpark you'll need an or map over RDD. Let's use the first option. First a couple of imports:

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

和一个函数:

as_ml = udf(lambda v: v.asML() if v is not None else None, VectorUDT())

使用示例数据:

from pyspark.mllib.linalg import Vectors as MLLibVectors

df = sc.parallelize([
    (MLLibVectors.sparse(4, [0, 2], [1, -1]), ),
    (MLLibVectors.dense([1, 2, 3, 4]), )
]).toDF(["features"])

result = df.withColumn("features", as_ml("features"))

结果是

+--------------------+
|            features|
+--------------------+
|(4,[0,2],[1.0,-1.0])|
|   [1.0,2.0,3.0,4.0]|
+--------------------+

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