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python – 将ML VectorUDT功能从.mllib转换为.ml类型进行线性回

发布时间:2020-12-16 22:40:33 所属栏目:Python 来源:网络整理
导读:我正在使用Spark集群,我想通过执行此代码来实现线性回归: data = sqlContext.read.format("com.databricks.spark.csv") .option("header","true") .option("inferSchema","true") .load("/FileStore/tables/w4s3yhez1497323663423/basma.csv/")data.c

我正在使用Spark集群,我想通过执行此代码来实现线性回归:

data = sqlContext.read.format("com.databricks.spark.csv") 
    .option("header","true") 
    .option("inferSchema","true") 
    .load("/FileStore/tables/w4s3yhez1497323663423/basma.csv/")

data.cache()  # Cache data for faster reuse
data.count()

from pyspark.mllib.regression import LabeledPoint

# convenience for specifying schema
data = data.select("creat0","gfr0m").rdd.map(lambda r: LabeledPoint(r[1],[r[0]])) 
    .toDF()
display(data)

from pyspark.ml.feature import VectorAssembler

vecAssembler = VectorAssembler(inputCols=["creat0","gfr0m"],outputCol="features")

(trainingData,testData) = data.randomSplit([0.7,0.3],seed=100)

trainingData.cache()
testData.cache()

print "Training Data : ",trainingData.count()
print "Test Data : ",testData.count()

data.collect()

from pyspark.ml.regression import LinearRegression

lr = LinearRegression()
# Fit 2 models,using different regularization parameters
modelA = lr.fit(data,{lr.regParam: 0.0})
modelB = lr.fit(data,{lr.regParam: 100.0})


# Make predictions
 predictionsA = modelA.transform(data)
  display(predictionsA)
   from pyspark.ml.evaluation import RegressionEvaluator
  evaluator = RegressionEvaluator(metricName="rmse")
  RMSE = evaluator.evaluate(predictionsA)
 print("ModelA: Root Mean Squared Error = " + str(RMSE))

# ModelA: Root Mean Squared Error = 128.602026843
 predictionsB = modelB.transform(data)
 RMSE = evaluator.evaluate(predictionsB)
  print("ModelB: Root Mean Squared Error = " + str(RMSE))

     # ModelB: Root Mean Squared Error = 129.496300193
  # Import numpy,pandas,and ggplot
  import numpy as np
 from pandas import *
   from ggplot import *


 But its give me this error:

IllegalArgumentException: u’requirement failed: Column features must
be of type org.apache.spark.ml.linalg.VectorUDT@3bfc3ba7 but was
actually org.apache.spark.mllib.linalg.VectorUDT@f71b0bce.

谷歌搜索此错误后,我找到了一个答案:

使用pyspark.ml.linalg import Vectors,VectorUDT

代替

from pyspark.mllib.linalg import Vectors,VectorUDT 

要么

from pyspark.ml.linalg import VectorUDT
from pyspark.sql.functions import udf

和功能:

  as_ml = udf(lambda v: v.asML() if v is not None else None,VectorUDT())

使用示例数据:

from pyspark.mllib.linalg import Vectors as MLLibVectors

df = sc.parallelize([
    (MLLibVectors.sparse(4,[0,2],[1,-1]),),(MLLibVectors.dense([1,2,3,4]),)
]).toDF(["features"])

result = df.withColumn("features",as_ml("features"))

但我仍然有同样的错误:

这里有一些数据:

原因,“weight0”,“dbp0”,“gfr0m”
1,“90”,“10”,“22.72”
5,“54”,“16.08”
6,“66”,“9”,“25.47”
3,“110”,“11”,“32.95”
5,“62”,“20.3”
5,“65”,“8”,“28.94”
1,“15.88”
5,“96”,“38.09”
5,“41.64”
如图4所示,“68”,“25.85”
5,“7”,“37.77”
1,“82”,“9.5”,“16.25”
5,“76”,“37.55”
5,“56”,“”,“37.06”
1,“93”,“18.26”
5,“80”,“7.5”,“48.49”
1,“73”,“38.37”
如图4所示,“31.09”
1,“39.62”
1,“40.08”
1,“28.2”
5,“81”,“36.66”
2,“47.1”
5,“91”,“16.59”
2,“58”,“49.22”
1,“38.98”
,“61”,“21.8”
5,“50”,“6”,“26.97”
1,“83”,“27.81”
1,“86”,“48.62”
,“77”,“46.78”
5,“64”,“34.17”
5,“38.95”
1,“7.63”
5,“32.46”
1,“35.98”
5,“32.26”
5,“42”,“17.3”
1,“88”,“25.61”
5,“”
1,“84”,“31.4”
5,“53.25”
1,“52.65”
6,“74”,“40.77”
1,“70”,“22.35”
6,“20.16”
1,“52”,“13”,“32.61”
,“52.98”
5,“28.67”

最佳答案
在这里,您只需要从pyspark.ml中为VectorUDT设置别名:

from pyspark.mllib.linalg import Vectors as MLLibVectors
from pyspark.ml.linalg import VectorUDT as VectorUDTML
from pyspark.sql.functions import udf

as_ml = udf(lambda v: v.asML() if v is not None else None,VectorUDTML())

df = sc.parallelize([
    (MLLibVectors.sparse(4,as_ml("features"))

result.show()
#?+--------------------+
# |            features|
#?+--------------------+
# |(4,[1.0,-1.0])|
# |   [1.0,2.0,3.0,4.0]|
# +--------------------+

当然,生成的DataFrame结果还没有准备好传递给LinearRegression,因为它没有标签列,但我相信你知道如何处理它.

(编辑:李大同)

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