LogisticRegressionModel 手动预测

Question

我试图预测 DataFrame 中每一行的标签，但没有使用LinearRegressionModel 的transform 方法，由于别有用心，我尝试使用经典公式 1 / (1 + e^(-hθ(x))) 手动计算它，请注意，我从 Apache Spark 的存储库中复制了代码，并从 private 中复制了几乎所有内容object BLAS 变成它的 public 版本。 PD：我没有使用任何regParam，我只是拟合模型。

//Notice that I had to obtain intercept, and coefficients from my model
val intercept = model.intercept
val coefficients = model.coefficients

val margin: Vector => Double = (features) => {
  BLAS.dot(features, coefficients) + intercept
}

val score: Vector => Double = (features) => {
  val m = margin(features)
  1.0 / (1.0 + math.exp(-m))
}

定义这些函数并获取模型参数后，我创建了一个 UDF 来计算预测（它接收与 DenseVector 相同的特征），稍后我将我的预测与真实模型的预测进行比较而且它们 非常不同！ 那么我错过了什么？我做错了什么？

val predict = udf((v: DenseVector) => {
  val recency = v(0)
  val frequency = v(1)
  val tp = score(new DenseVector(Array(recency, frequency)))
  new DenseVector(Array(tp, 1 - tp))
})

// model's predictions
val xf = model.transform(df)

df.select(col("id"), predict(col("features")).as("myprediction"))
  .join(xf, df("id") === xf("id"), "inner")
  .select(df("id"), col("probability"), col("myprediction"))
  .show

+----+--------------------+--------------------+
|  id|         probability|        myprediction|
+----+--------------------+--------------------+
|  31|[0.97579780436514...|[0.98855386037790...|
| 231|[0.97579780436514...|[0.98855386037790...|
| 431|[0.69794428333266...|           [1.0,0.0]|
| 631|[0.97579780436514...|[0.98855386037790...|
| 831|[0.97579780436514...|[0.98855386037790...|
|1031|[0.96509616791398...|[0.99917463322937...|
|1231|[0.96509616791398...|[0.99917463322937...|
|1431|[0.96509616791398...|[0.99917463322937...|
|1631|[0.94231815700848...|[0.99999999999999...|
|1831|[0.96509616791398...|[0.99917463322937...|
|2031|[0.96509616791398...|[0.99917463322937...|
|2231|[0.96509616791398...|[0.99917463322937...|
|2431|[0.95353743438055...|           [1.0,0.0]|
|2631|[0.94646924057674...|           [1.0,0.0]|
|2831|[0.96509616791398...|[0.99917463322937...|
|3031|[0.96509616791398...|[0.99917463322937...|
|3231|[0.95971207153567...|[0.99999999999996...|
|3431|[0.96509616791398...|[0.99917463322937...|
|3631|[0.96509616791398...|[0.99917463322937...|
|3831|[0.96509616791398...|[0.99917463322937...|
+----+--------------------+--------------------+

编辑

我什至尝试在 udf 中定义此类函数，但没有成功。

def predict(coefficients: Vector, intercept: Double) = {
  udf((v: DenseVector) => {
    def margin(features: Vector, coefficients: Vector, intercept: Double): Double = {
      BLAS.dot(features, coefficients) + intercept
    }

    def score(features: Vector, coefficients: Vector, intercept: Double): Double = {
      val m = margin(features, coefficients, intercept)
      1.0 / (1.0 + math.exp(-m))
    }

    val recency = v(0)
    val frequency = v(1)
    val tp = score(new DenseVector(Array(recency, frequency)), coefficients, intercept)
    new DenseVector(Array(tp, 1 - tp))
  })
}

Answer 1

很尴尬，其实问题是因为我用了一个Pipeline，加了一个MinMaxScaler作为stage，所以数据集在模型训练之前就被缩放了，所以两个参数coefficients 和 intercept 与 缩放数据 相关联，因此当我使用它们计算预测时，结果完全有偏差。因此，为了解决这个问题，我只是对训练数据集进行了非标准化，这样我就可以获得那些 coefficients 和 intercept。重新执行代码后，我得到了与Spark相同的结果。另一方面，我听取了 @zero323 的建议并将 margin 和 score 定义移到了 udf 的第一个 [=20] 中=]声明。

LogisticRegressionModel 手动预测

LogisticRegressionModel prediction manually

scala

logistic-regression

apache-spark

编辑