Apache spark Pyspark的交叉验证度量
当我们进行k倍交叉验证时,我们正在测试一个模型在预测从未见过的数据时表现得有多好 如果将我的数据集分成90%的训练和10%的测试,并分析模型性能,则无法保证我的测试集不包含10%的“最容易”或“最难”预测点 通过进行10倍交叉验证,我可以确保每个点至少用于一次培训。由于(在本例中)将对模型进行10次测试,我们可以对这些测试指标进行分析,这将使我们更好地了解模型在分类新数据方面的表现 交叉验证是一种优化算法超参数的方法,其目的是检查模型 通过这样做:Apache spark Pyspark的交叉验证度量,apache-spark,pyspark,apache-spark-mllib,cross-validation,Apache Spark,Pyspark,Apache Spark Mllib,Cross Validation,当我们进行k倍交叉验证时,我们正在测试一个模型在预测从未见过的数据时表现得有多好 如果将我的数据集分成90%的训练和10%的测试,并分析模型性能,则无法保证我的测试集不包含10%的“最容易”或“最难”预测点 通过进行10倍交叉验证,我可以确保每个点至少用于一次培训。由于(在本例中)将对模型进行10次测试,我们可以对这些测试指标进行分析,这将使我们更好地了解模型在分类新数据方面的表现 交叉验证是一种优化算法超参数的方法,其目的是检查模型 通过这样做: lr = LogisticRegression
lr = LogisticRegression(maxIter=10, tol=1E-4)
ovr = OneVsRest(classifier=lr)
pipeline = Pipeline(stages=[... , ovr])
crossval = CrossValidator(estimator=pipeline,
estimatorParamMaps=paramGrid,
evaluator=MulticlassClassificationEvaluator(),
numFolds=10)
# Run cross-validation, and choose the best set of parameters.
cvModel = crossval.fit(df)
据我所知,我能够获得一个模型,该模型具有在paramGrid中定义的最佳参数集。我理解这种超参数调整的价值,但我想要的是分析模型性能,而不仅仅是获得最佳模型
问题是(对于这种情况下的10倍交叉验证):
是否可以使用CrossValidator为10个测试中的每一个(或每个度量的10个测试的平均值)提取度量(f1、精度、召回率等)?即,是否可以使用CrossValidator进行模型检查而不是模型选择
谢谢
更新
如评论中所述,可以找到类似的问题。第一个建议是在拟合之前将collectSubModels设置为true,这会抛出一个错误,表示关键字不存在(老实说,我没有花很多时间试图找出原因) 用户在其回答中提供了一种打印中间培训结果的变通方法。使用他提供的方法,可以打印评估指标的中间结果。由于我想提取精度、召回率、f1和混淆矩阵的中间结果,我对他实施的方法做了一些修改:
TestResult = collections.namedtuple("TestResult", ["params", "metrics"])
class CrossValidatorVerbose(CrossValidator):
def _fit(self, dataset):
folds = []
est = self.getOrDefault(self.estimator)
epm = self.getOrDefault(self.estimatorParamMaps)
numModels = len(epm)
eva = self.getOrDefault(self.evaluator)
metricName = eva.getMetricName()
nFolds = self.getOrDefault(self.numFolds)
seed = self.getOrDefault(self.seed)
h = 1.0 / nFolds
randCol = self.uid + "_rand"
df = dataset.select("*", rand(seed).alias(randCol))
metrics = [0.0] * numModels
for i in range(nFolds):
folds.append([])
foldNum = i + 1
print("Comparing models on fold %d" % foldNum)
validateLB = i * h
validateUB = (i + 1) * h
condition = (df[randCol] >= validateLB) & (df[randCol] < validateUB)
validation = df.filter(condition)
train = df.filter(~condition)
for j in range(numModels):
paramMap = epm[j]
model = est.fit(train, paramMap)
# TODO: duplicate evaluator to take extra params from input
prediction = model.transform(validation, paramMap)
metric = eva.evaluate(prediction)
metrics[j] += metric
avgSoFar = metrics[j] / foldNum
print("params: %s\t%s: %f\tavg: %f" % (
{param.name: val for (param, val) in paramMap.items()},
metricName, metric, avgSoFar))
predictionLabels = prediction.select("prediction", "label")
allMetrics = MulticlassMetrics(predictionLabels.rdd)
folds[i].append(TestResult(paramMap.items(), allMetrics))
if eva.isLargerBetter():
bestIndex = np.argmax(metrics)
else:
bestIndex = np.argmin(metrics)
bestParams = epm[bestIndex]
bestModel = est.fit(dataset, bestParams)
avgMetrics = [m / nFolds for m in metrics]
bestAvg = avgMetrics[bestIndex]
print("Best model:\nparams: %s\t%s: %f" % (
{param.name: val for (param, val) in bestParams.items()},
metricName, bestAvg))
return self._copyValues(CrossValidatorModel(bestModel, avgMetrics)), folds
要打印特定折叠的度量(使用第一组超参数打印第一个折叠):
将打印如下内容:
Class 0.0 precision = 0.809523809524
Class 0.0 recall = 0.772727272727
Class 0.0 F1 Measure = 0.790697674419
Class 1.0 precision = 0.857142857143
Class 1.0 recall = 0.818181818182
Class 1.0 F1 Measure = 0.837209302326
Class 2.0 precision = 0.875
Class 2.0 recall = 0.875
Class 2.0 F1 Measure = 0.875
...
Weighted recall = 0.808333333333
Weighted precision = 0.812411616162
Weighted F(1) Score = 0.808461689698
Weighted F(0.5) Score = 0.810428077222
Weighted false positive rate = 0.026335560185
Accuracy = 0.808333333333
您还可以检查可能的副本-其中一个答案可能对您有所帮助。
def printMetrics(metrics, df):
labels = df.rdd.map(lambda lp: lp.label).distinct().collect()
for label in sorted(labels):
print("Class %s precision = %s" % (label, metrics.precision(label)))
print("Class %s recall = %s" % (label, metrics.recall(label)))
print("Class %s F1 Measure = %s" % (label, metrics.fMeasure(label, beta=1.0)))
print ""
# Weighted stats
print("Weighted recall = %s" % metrics.weightedRecall)
print("Weighted precision = %s" % metrics.weightedPrecision)
print("Weighted F(1) Score = %s" % metrics.weightedFMeasure())
print("Weighted F(0.5) Score = %s" % metrics.weightedFMeasure(beta=0.5))
print("Weighted false positive rate = %s" % metrics.weightedFalsePositiveRate)
print("Accuracy = %s" % metrics.accuracy)
printMetrics(folds[0][0].metrics, df)
Class 0.0 precision = 0.809523809524
Class 0.0 recall = 0.772727272727
Class 0.0 F1 Measure = 0.790697674419
Class 1.0 precision = 0.857142857143
Class 1.0 recall = 0.818181818182
Class 1.0 F1 Measure = 0.837209302326
Class 2.0 precision = 0.875
Class 2.0 recall = 0.875
Class 2.0 F1 Measure = 0.875
...
Weighted recall = 0.808333333333
Weighted precision = 0.812411616162
Weighted F(1) Score = 0.808461689698
Weighted F(0.5) Score = 0.810428077222
Weighted false positive rate = 0.026335560185
Accuracy = 0.808333333333