我正在尝试使用PySpark对样本数据进行逻辑回归。面对申请' LabeledPoint'哈希之后。
输入数据框:
+--+--------+
|C1| C2|
+--+--------+
| 0|776ce399|
| 0|3486227d|
| 0|e5ba7672|
| 1|3486227d|
| 0|e5ba7672|
+--+--------+
在C2列上应用散列后,
tokenizer = Tokenizer(inputCol="C2", outputCol="words")
wordsData = tokenizer.transform(df)
hashingTF = HashingTF(inputCol="words", outputCol="rawFeatures", numFeatures=20)
featurizedData = hashingTF.transform(wordsData)
idf = IDF(inputCol="rawFeatures", outputCol="features")
idfModel = idf.fit(featurizedData)
rescaledData = idfModel.transform(featurizedData)
+--+--------+--------------------+---------------+--------------------+
|C1| C2| words| rawFeatures| features|
+--+--------+--------------------+---------------+--------------------+
| 0|776ce399|ArrayBuffer(776ce...|(20,[15],[1.0])|(20,[15],[2.30003...|
| 0|3486227d|ArrayBuffer(34862...| (20,[0],[1.0])|(20,[0],[2.455603...|
| 0|e5ba7672|ArrayBuffer(e5ba7...| (20,[9],[1.0])|(20,[9],[0.660549...|
| 1|3486227d|ArrayBuffer(34862...| (20,[0],[1.0])|(20,[0],[2.455603...|
| 0|e5ba7672|ArrayBuffer(e5ba7...| (20,[9],[1.0])|(20,[9],[0.660549...|
+--+--------+--------------------+---------------+--------------------+
现在应用逻辑回归,当我执行LabeledPoint
时 temp = rescaledData.map(lambda line: LabeledPoint(line[0],line[4]))
得到以下错误,
ValueError: setting an array element with a sequence.
请帮忙。
答案 0 :(得分:0)
感谢您提出建议零点。
使用管道概念实现。
from pyspark.ml import Pipeline
from pyspark.ml.classification import LogisticRegression
from pyspark.ml.feature import HashingTF, Tokenizer, IDF
from pyspark.sql import Row
from pyspark.sql.functions import col
from pyspark.sql.types import DoubleType
dfWithLabel = df.withColumn("label", col("C1").cast(DoubleType()))
tokenizer = Tokenizer(inputCol="C2", outputCol="D2")
hashingTF = HashingTF(inputCol=tokenizer.getOutputCol(), outputCol="E2")
idf = IDF(inputCol=hashingTF.getOutputCol(), outputCol="features")
lr = LogisticRegression(maxIter=10, regParam=0.01)
pipeline = Pipeline(stages=[tokenizer, hashingTF,idf,lr])
# Fit the pipeline to training documents.
model = pipeline.fit(dfWithLabel)