通过pyspark.ml CrossValidator调整隐式pyspark.ml ALS矩阵分解模型的参数

Question

我正在尝试调整使用隐式数据的ALS矩阵分解模型的参数.为此,我正在尝试使用pyspark.ml.tuning.CrossValidator来运行参数网格并选择最佳模型.我相信我的问题在于评估者,但我无法弄明白.

我可以使用回归RMSE评估器为显式数据模型工作,如下所示:

from pyspark import SparkConf, SparkContext
from pyspark.sql import SQLContext
from pyspark.ml.recommendation import ALS
from pyspark.ml.tuning import CrossValidator, ParamGridBuilder
from pyspark.ml.evaluation import BinaryClassificationEvaluator
from pyspark.ml.evaluation import RegressionEvaluator

from pyspark.sql.functions import rand


conf = SparkConf() \
  .setAppName("MovieLensALS") \
  .set("spark.executor.memory", "2g")
sc = SparkContext(conf=conf)

sqlContext = SQLContext(sc)

dfRatings = sqlContext.createDataFrame([(0, 0, 4.0), (0, 1, 2.0), (1, 1, 3.0), (1, 2, 4.0), (2, 1, 1.0), (2, 2, 5.0)],
                                 ["user", "item", "rating"])
dfRatingsTest = sqlContext.createDataFrame([(0, 0), (0, 1), (1, 1), (1, 2), (2, 1), (2, 2)], ["user", "item"])

alsExplicit = ALS()
defaultModel = alsExplicit.fit(dfRatings)

paramMapExplicit = ParamGridBuilder() \
                    .addGrid(alsExplicit.rank, [8, 12]) \
                    .addGrid(alsExplicit.maxIter, [10, 15]) \
                    .addGrid(alsExplicit.regParam, [1.0, 10.0]) \
                    .build()

evaluatorR = RegressionEvaluator(metricName="rmse", labelCol="rating")

cvExplicit = CrossValidator(estimator=alsExplicit, estimatorParamMaps=paramMapExplicit, evaluator=evaluatorR)
cvModelExplicit = cvExplicit.fit(dfRatings)

predsExplicit = cvModelExplicit.bestModel.transform(dfRatingsTest)
predsExplicit.show()

当我尝试为隐式数据执行此操作时(假设视图计数而不是评级),我得到一个我无法弄清楚的错误.这是代码(与上面非常相似):

dfCounts = sqlContext.createDataFrame([(0,0,0), (0,1,12), (0,2,3), (1,0,5), (1,1,9), (1,2,0), (2,0,0), (2,1,11), (2,2,25)],
                                 ["user", "item", "rating"])
dfCountsTest = sqlContext.createDataFrame([(0, 0), (0, 1), (1, 1), (1, 2), (2, 1), (2, 2)], ["user", "item"])

alsImplicit = ALS(implicitPrefs=True)
defaultModelImplicit = alsImplicit.fit(dfCounts)

paramMapImplicit = ParamGridBuilder() \
                    .addGrid(alsImplicit.rank, [8, 12]) \
                    .addGrid(alsImplicit.maxIter, [10, 15]) \
                    .addGrid(alsImplicit.regParam, [1.0, 10.0]) \
                    .addGrid(alsImplicit.alpha, [2.0,3.0]) \
                    .build()

evaluatorB = BinaryClassificationEvaluator(metricName="areaUnderROC", labelCol="rating")
evaluatorR = RegressionEvaluator(metricName="rmse", labelCol="rating")

cv = CrossValidator(estimator=alsImplicit, estimatorParamMaps=paramMapImplicit, evaluator=evaluatorR)
cvModel = cv.fit(dfCounts)

predsImplicit = cvModel.bestModel.transform(dfCountsTest)
predsImplicit.show()

我尝试使用RMSE评估程序执行此操作,但是出现错误.据我所知,我还应该能够将AUC度量用于二​​元分类评估器,因为隐式矩阵分解的预测是用于预测二进制矩阵p_ui的置信矩阵c_ui,本文为pyspark ALS的文档.引用.

使用评估器给我一个错误,我找不到任何有关在线交叉验证隐式ALS模型的富有成效的讨论.我正在查看CrossValidator源代码,试图弄清楚出了什么问题,但遇到了麻烦.我的一个想法是,在该过程将隐式数据矩阵r_ui转换为二进制矩阵p_ui和置信矩阵c_ui之后,我不确定它在评估阶段比较预测的c_ui矩阵是什么.

这是错误:

Traceback (most recent call last):

  File "<ipython-input-16-6c43b997005e>", line 1, in <module>
    cvModel = cv.fit(dfCounts)

  File "C:/spark-1.6.1-bin-hadoop2.6/python\pyspark\ml\pipeline.py", line 69, in fit
    return self._fit(dataset)

  File "C:/spark-1.6.1-bin-hadoop2.6/python\pyspark\ml\tuning.py", line 239, in _fit
    model = est.fit(train, epm[j])

  File "C:/spark-1.6.1-bin-hadoop2.6/python\pyspark\ml\pipeline.py", line 67, in fit
    return self.copy(params)._fit(dataset)

  File "C:/spark-1.6.1-bin-hadoop2.6/python\pyspark\ml\wrapper.py", line 133, in _fit
    java_model = self._fit_java(dataset)

  File "C:/spark-1.6.1-bin-hadoop2.6/python\pyspark\ml\wrapper.py", line 130, in _fit_java
    return self._java_obj.fit(dataset._jdf)

  File "C:\spark-1.6.1-bin-hadoop2.6\python\lib\py4j-0.9-src.zip\py4j\java_gateway.py", line 813, in __call__
    answer, self.gateway_client, self.target_id, self.name)

  File "C:/spark-1.6.1-bin-hadoop2.6/python\pyspark\sql\utils.py", line 45, in deco
    return f(*a, **kw)

  File "C:\spark-1.6.1-bin-hadoop2.6\python\lib\py4j-0.9-src.zip\py4j\protocol.py", line 308, in get_return_value
    format(target_id, ".", name), value)

etc.......

UPDATE

我尝试缩放输入,使其在0到1的范围内并使用RMSE评估器.它似乎工作得很好,直到我尝试将其插入CrossValidator.

以下代码有效.我得到了预测,我从评估员那里得到了一个RMSE值.

from pyspark import SparkConf, SparkContext
from pyspark.sql import SQLContext
from pyspark.sql.types import FloatType
import pyspark.sql.functions as F
from pyspark.ml.recommendation import ALS
from pyspark.ml.tuning import CrossValidator, ParamGridBuilder
from pyspark.ml.evaluation import RegressionEvaluator


conf = SparkConf() \
  .setAppName("ALSPractice") \
  .set("spark.executor.memory", "2g")
sc = SparkContext(conf=conf)

sqlContext = SQLContext(sc)

# Users 0, 1, 2, 3 - Items 0, 1, 2, 3, 4, 5 - Ratings 0.0-5.0
dfCounts2 = sqlContext.createDataFrame([(0,0,5.0), (0,1,5.0),            (0,3,0.0), (0,4,0.0), 
                                        (1,0,5.0),            (1,2,4.0), (1,3,0.0), (1,4,0.0),
                                        (2,0,0.0),            (2,2,0.0), (2,3,5.0), (2,4,5.0),
                                        (3,0,0.0), (3,1,0.0),            (3,3,4.0)            ],
                                       ["user", "item", "rating"])

dfCountsTest2 = sqlContext.createDataFrame([(0,0), (0,1), (0,2), (0,3), (0,4),
                                            (1,0), (1,1), (1,2), (1,3), (1,4),
                                            (2,0), (2,1), (2,2), (2,3), (2,4),
                                            (3,0), (3,1), (3,2), (3,3), (3,4)], ["user", "item"])

# Normalize rating data to [0,1] range based on max rating
colmax = dfCounts2.select(F.max('rating')).collect()[0].asDict().values()[0]
normalize = udf(lambda x: x/colmax, FloatType())
dfCountsNorm = dfCounts2.withColumn('ratingNorm', normalize(col('rating')))

alsImplicit = ALS(implicitPrefs=True)
defaultModelImplicit = alsImplicit.fit(dfCountsNorm)
preds = defaultModelImplicit.transform(dfCountsTest2)

evaluatorR2 = RegressionEvaluator(metricName="rmse", labelCol="ratingNorm")
evaluatorR2.evaluate(defaultModelImplicit.transform(dfCountsNorm))

preds = defaultModelImplicit.transform(dfCountsTest2)

我不明白的是为什么以下不起作用.我使用相同的估算器,相同的评估器并拟合相同的数据.为什么这些工作在上面但不在CrossValidator中:

paramMapImplicit = ParamGridBuilder() \
                    .addGrid(alsImplicit.rank, [8, 12]) \
                    .addGrid(alsImplicit.maxIter, [10, 15]) \
                    .addGrid(alsImplicit.regParam, [1.0, 10.0]) \
                    .addGrid(alsImplicit.alpha, [2.0,3.0]) \
                    .build()

cv = CrossValidator(estimator=alsImplicit, estimatorParamMaps=paramMapImplicit, evaluator=evaluatorR2)
cvModel = cv.fit(dfCountsNorm)

Answer 1

忽略技术问题,严格来说,鉴于ALS使用隐式反馈生成的输入,两种方法都不正确.

• 您无法使用RegressionEvaluator,因为您已经知道,预测可以被解释为置信度值,并表示为范围[0,1]中的浮点数,而标签列只是一个未绑定的整数.这些值显然无法比较.
• 你不能使用BinaryClassificationEvaluator因为即使预测可以解释为概率标签也不代表二元决策.此外,预测列具有无效类型,不能直接使用BinaryClassificationEvaluator

您可以尝试转换其中一列,使输入符合要求,但从理论角度来看,这并不是一种合理的方法,并引入了难以调整的其他参数.

• 将标签列映射到[0,1]范围并使用RMSE.

• 将标签列转换为具有固定阈值的二进制指示符,并扩展ALS/ ALSModel以返回预期的列类型.假设阈值为1,则可能是这样的

  from pyspark.ml.recommendation import *
  from pyspark.sql.functions import udf, col
  from pyspark.mllib.linalg import DenseVector, VectorUDT
  
  class BinaryALS(ALS):
      def fit(self, df):
          assert self.getImplicitPrefs()
          model = super(BinaryALS, self).fit(df)
          return ALSBinaryModel(model._java_obj)
  
  class ALSBinaryModel(ALSModel):
      def transform(self, df):
          transformed = super(ALSBinaryModel, self).transform(df)
          as_vector = udf(lambda x: DenseVector([1 - x, x]), VectorUDT())
          return transformed.withColumn(
              "rawPrediction", as_vector(col("prediction")))
  
  # Add binary label column
  with_binary = dfCounts.withColumn(
      "label_binary", (col("rating") > 0).cast("double"))
  
  als_binary_model = BinaryALS(implicitPrefs=True).fit(with_binary)
  
  evaluatorB = BinaryClassificationEvaluator(
      metricName="areaUnderROC", labelCol="label_binary")
  
  evaluatorB.evaluate(als_binary_model.transform(with_binary))
  ## 1.0
  

一般来说,关于评估具有隐式反馈的推荐系统的材料在教科书中是缺失的,我建议你阅读一下eliasah关于评估这类推荐者的答案.