我正在使用Spark 1.5.1和 在pyspark中,在我使用:
拟合模型之后model = LogisticRegressionWithLBFGS.train(parsedData)
我可以使用以下方式打印预测:
model.predict(p.features)
是否还有一个功能来打印概率分数以及预测?
答案 0 :(得分:7)
您必须先clear the threshold,这仅适用于二进制分类:
from pyspark.mllib.classification import LogisticRegressionWithLBFGS, LogisticRegressionModel
from pyspark.mllib.regression import LabeledPoint
parsed_data = [LabeledPoint(0.0, [4.6,3.6,1.0,0.2]),
LabeledPoint(0.0, [5.7,4.4,1.5,0.4]),
LabeledPoint(1.0, [6.7,3.1,4.4,1.4]),
LabeledPoint(0.0, [4.8,3.4,1.6,0.2]),
LabeledPoint(1.0, [4.4,3.2,1.3,0.2])]
model = LogisticRegressionWithLBFGS.train(sc.parallelize(parsed_data))
model.threshold
# 0.5
model.predict(parsed_data[2].features)
# 1
model.clearThreshold()
model.predict(parsed_data[2].features)
# 0.9873840020002339
答案 1 :(得分:0)
我认为问题在于计算预测整个训练集的概率分数。如果是这样,我做了以下计算。不确定帖子是否仍然有效,但这就是我这样做的方式:
#get the original training data before it was converted to rows of LabelPoint.
#let us assume it is otd ( of type spark DataFrame)
#let us extract the featureset as rdd by:
fs=otd.rdd.map(lambda x:x[1:]) # assuming label is col 0.
#the below is just a sample way of creating a Labelpoint rows..
parsedData= otd.rdd.map(lambda x: reg.LabeledPoint(int(x[0]-1),x[1:]))
# now convert otd to a panda DataFrame as:
ptd= otd.toPandas()
m= ptd.shape[0]
# train and get the model
model=LogisticRegressionWithLBFGS.train(trainingData,numClasses=10)
#Now store the model.predict rdd structures
predict=model.predict(fs)
pr=predict.collect()
correct=0
correct = ((ptd.label-1) == (pr)).sum()
print((correct/m) *100)
注意以上是针对多级分类的。