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python - 使用 spark ml(数据框)进行逻辑回归

转载 作者:太空宇宙 更新时间:2023-11-04 08:02:44 25 4
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我为逻辑回归写了下面的代码,我想使用spark.ml提供的pipeline API。但是,在我尝试打印系数和截距后,它给了我一个错误。此外,我在计算混淆矩阵和其他指标(如精度、召回率)时遇到了问题。

#Logistic Regression:
from pyspark.mllib.linalg import Vectors
from pyspark.ml.classification import LogisticRegression
from pyspark.sql import SQLContext
from pyspark import SparkContext
from pyspark.sql.types import *
from pyspark.sql.functions import *
from pyspark.ml.feature import StringIndexer,VectorAssembler
from pyspark.ml import Pipeline
from pyspark.ml.evaluation import MulticlassClassificationEvaluator


sc = SparkContext("local", "predictive")
sqlContext=SQLContext(sc)

df = sqlContext.read.load('/user/bna_ads_final.csv',
format='com.databricks.spark.csv',
header='true',
inferSchema='true')

df.show(5)
df.count()
df.dtypes
df=df.withColumn("load_date",df.load_date.cast("timestamp"))
df_withday= df.withColumn("day",dayofmonth(df.load_date))
df_new=df_withday.withColumn("Month",month(df.load_date))
df_new=df_new.withColumn("classname",df_new.classname.cast("string"))
ignore = ["load_date","wo_flag","serialnumber", "classname"]

def modify_values(r):
if r == "A" or r =="B":
return "dispatch"
else:
return "non-dispatch"

def show_metrics(metrics):
# Overall statistics
precision = metrics.precision()
recall = metrics.recall()
f1Score = metrics.fMeasure()
print("Summary Stats")
print("Precision = %s" % precision)
print("Recall = %s" % recall)
print("F1 Score = %s" % f1Score)
print (metrics.confusionMatrix())

ol_val = udf(modify_values, StringType())
df_final = df_new.withColumn("wo_flag",ol_val(df_new.wo_flag))
indexer= StringIndexer(inputCol="classname", outputCol="classnamecat")
indexed = indexer.fit(df_final).transform(df_final)
indexed=indexed.withColumn("classnamecat",indexed.classnamecat.cast("int"))
indexed.show(5)
(trainingData, testData) = indexed.randomSplit([0.7, 0.3])
assembler = VectorAssembler(inputCols=[x for x in indexed.columns if x not in ignore],outputCol='features')
stringindexer=StringIndexer(inputCol="wo_flag", outputCol="labellr")
Classifier= LogisticRegression(labelCol="labellr", featuresCol="features")
pipeline=Pipeline(stages=[stringindexer,assembler,Classifier])
model = pipeline.fit(trainingData)
predictions = model.transform(testData)

selected = predictions.select("features", "labellr", "probability", "prediction")
for row in selected.collect():
print row


evaluator = MulticlassClassificationEvaluator(
labelCol="labellr", predictionCol="prediction", metricName="precision")
accuracy = evaluator.evaluate(predictions)
print("Test Error = %g" % (1.0 - accuracy))
print("Accuracy= %g" % (accuracy))

print("Coefficients: " + str(model.coefficients))
print("Intercept: " + str(model.intercept))

我得到的错误是:

print("Coefficients: " + str(model.coefficients))
AttributeError: 'PipelineModel' object has no attribute 'coefficients'

我在 Hadoop 集群上安装了 Spark 1.5,短期内无法升级。是否有解决此问题的变通方法。

load_date           |  r         |   classname| mstatus34_timdiff|  day|Month| classnamecat| serialnumber
+-----------+------------------+----------+--------------------+------------+--- +-----------+----
2013-12-29 10:55:...|non-dispatch| 6634| 19| 1| 7| 0.0| 231234
2014-10-05 23:43:...|non-dispatch| 6634| 4| 5| 10| 0.0| 342345
2014-10-09 09:39:...| dispatch| 5886| 36| 9| 10| 1.0| 563472
2014-09-16 09:47:...| dispatch| 6634| 53| 16| 9| 0.0| 134657

最佳答案

您可以使用 PipelineModelstages 属性访问各个阶段

from pyspark.ml import Pipeline
from pyspark.ml.classification import LogisticRegression, LogisticRegressionModel
from pyspark.ml.feature import VectorAssembler

df = sc.parallelize([
(0.0, 1.0, 2.0, 4.0),
(1.0, 3.0, 4.0, 5.0)
]).toDF(["label", "x1", "x2", "x3"])

assembler = (VectorAssembler()
.setInputCols(df.columns[1:])
.setOutputCol("features"))

lr = LogisticRegression(maxIter=10, regParam=0.01)

pipeline = Pipeline(stages=[assembler, lr])
model = pipeline.fit(data)

[stage.coefficients for stage in model.stages if hasattr(stage, "coefficients")]
## [DenseVector([2.1178, 1.6843, -1.8338])]

## or

[stage.coefficients for stage in model.stages
if isinstance(stage, LogisticRegressionModel)]
## [DenseVector([2.1178, 1.6843, -1.8338])]

关于python - 使用 spark ml(数据框)进行逻辑回归,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/37278999/

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