如何将 Spark Dense Matrix 转换为 Spark Dataframe

Question

我正在尝试在 Scala Spark 中实现一些代码，其中我有一个多类逻辑回归 Model 和 model 生成一个系数矩阵。

这是代码 -

val training = spark.read.format("libsvm").load("data/mllib/sample_multiclass_classification_data.txt")


training.show(false)
+-----+-----------------------------------------------------------+
|label|features                                                   |
+-----+-----------------------------------------------------------+
|1.0  |(4,[0,1,2,3],[-0.222222,0.5,-0.762712,-0.833333])          |
|1.0  |(4,[0,1,2,3],[-0.555556,0.25,-0.864407,-0.916667])         |
|1.0  |(4,[0,1,2,3],[-0.722222,-0.166667,-0.864407,-0.833333])    |
|1.0  |(4,[0,1,2,3],[-0.722222,0.166667,-0.694915,-0.916667])     |
|0.0  |(4,[0,1,2,3],[0.166667,-0.416667,0.457627,0.5])            |
|1.0  |(4,[0,2,3],[-0.833333,-0.864407,-0.916667])                |
|2.0  |(4,[0,1,2,3],[-1.32455E-7,-0.166667,0.220339,0.0833333])   |
|2.0  |(4,[0,1,2,3],[-1.32455E-7,-0.333333,0.0169491,-4.03573E-8])|
|1.0  |(4,[0,1,2,3],[-0.5,0.75,-0.830508,-1.0])                   |
|0.0  |(4,[0,2,3],[0.611111,0.694915,0.416667])                   |
|0.0  |(4,[0,1,2,3],[0.222222,-0.166667,0.423729,0.583333])       |
|1.0  |(4,[0,1,2,3],[-0.722222,-0.166667,-0.864407,-1.0])         |
|1.0  |(4,[0,1,2,3],[-0.5,0.166667,-0.864407,-0.916667])          |
|2.0  |(4,[0,1,2,3],[-0.222222,-0.333333,0.0508474,-4.03573E-8])  |
|2.0  |(4,[0,1,2,3],[-0.0555556,-0.833333,0.0169491,-0.25])       |
|2.0  |(4,[0,1,2,3],[-0.166667,-0.416667,-0.0169491,-0.0833333])  |
|1.0  |(4,[0,2,3],[-0.944444,-0.898305,-0.916667])                |
|2.0  |(4,[0,1,2,3],[-0.277778,-0.583333,-0.0169491,-0.166667])   |
|0.0  |(4,[0,1,2,3],[0.111111,-0.333333,0.38983,0.166667])        |
|2.0  |(4,[0,1,2,3],[-0.222222,-0.166667,0.0847457,-0.0833333])   |
+-----+-----------------------------------------------------------+

我正在尝试为 3 个标签安装 model。

scala> training.select("label").distinct.show
+-----+
|label|
+-----+
|  0.0|
|  1.0|
|  2.0|
+-----+

拟合逻辑回归 Model

import org.apache.spark.ml.classification.LogisticRegression
val lr = new LogisticRegression().setMaxIter(10).setRegParam(0.3).setElasticNetParam(0.8)
​
// Fit the model
val lrModel = lr.fit(training)
​

现在，当我尝试查看系数矩阵时，它给了我一个具有 3 行（用于 3 个标签）和 4 列（4 个输入特征）的矩阵

scala> lrModel.coefficientMatrix.toDense
res13: org.apache.spark.ml.linalg.DenseMatrix =
0.0  0.0  0.0                  0.3176483191238039
0.0  0.0  -0.7803943459681859  -0.3769611423403096
0.0  0.0  0.0                  0.0

这是每个标签的截距 -

scala> lrModel.interceptVector
res15: org.apache.spark.ml.linalg.Vector = [0.05165231659832854,-0.12391224990853622,0.07225993331020768]

我想使用系数矩阵和截距向量创建一个特征重要性Spark dataframe以获得最终结果 dataframe 像这样 -

label feature name  coefficient intercept
0         0             0         0.051
0         1             0         0.051
0         2             0         0.051
0         3             0.3176    0.051
1         0             0         -0.123
1         1             0         -0.123
1         2             -0.78     -0.123
1         3             -0.37     -0.123
2         0             0         0.072
2         1             0         0.072
2         2             0         0.072
2         3             0         0.072

每个特征对每个 label 都有一个系数，因此 output 中的总记录将是labels * features ，即3 * 4 = 12

我希望这个过程是动态的，将其包装在 function 中，以便我可以将其重新用于任意数量的功能和标签。

我正在从这里读取数据 - https://github.com/apache/spark/blob/master/data/mllib/sample_multiclass_classification_data.txt

Answer 1

我假设这里 lr 是以你的 pyspark 的逻辑回归为例。 下面的代码我尝试了多项逻辑回归

weights=lr.coefficientMatrix
rows = weights.toArray().tolist()
df = spark.createDataFrame(rows,["<your list of features columns used for training"])

上面的代码对我有用，不用担心序列分配你训练过的顺序