python - 如何将密集向量的RDD转换为pyspark中的DataFrame？

Question

我有一个像这样的DenseVectorRDD

>>> frequencyDenseVectors.collect()
[DenseVector([1.0, 0.0, 1.0, 1.0, 0.0, 0.0, 1.0, 0.0, 1.0, 1.0, 1.0, 0.0, 1.0]), DenseVector([1.0, 1.0, 1.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0, 0.0]), DenseVector([1.0, 1.0, 0.0, 1.0, 1.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0]), DenseVector([0.0, 1.0, 1.0, 0.0, 0.0, 1.0, 0.0, 0.0, 0.0, 0.0, 0.0, 1.0, 0.0])]

我想把它转换成Dataframe。我试过这样

>>> spark.createDataFrame(frequencyDenseVectors, ['rawfeatures']).collect()

它给出了这样的错误

Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "/opt/BIG-DATA/spark-2.0.0-bin-hadoop2.7/python/pyspark/sql/session.py", line 520, in createDataFrame
    rdd, schema = self._createFromRDD(data.map(prepare), schema, samplingRatio)
  File "/opt/BIG-DATA/spark-2.0.0-bin-hadoop2.7/python/pyspark/sql/session.py", line 360, in _createFromRDD
    struct = self._inferSchema(rdd, samplingRatio)
  File "/opt/BIG-DATA/spark-2.0.0-bin-hadoop2.7/python/pyspark/sql/session.py", line 340, in _inferSchema
    schema = _infer_schema(first)
  File "/opt/BIG-DATA/spark-2.0.0-bin-hadoop2.7/python/pyspark/sql/types.py", line 991, in _infer_schema
    fields = [StructField(k, _infer_type(v), True) for k, v in items]
  File "/opt/BIG-DATA/spark-2.0.0-bin-hadoop2.7/python/pyspark/sql/types.py", line 968, in _infer_type
    raise TypeError("not supported type: %s" % type(obj))
TypeError: not supported type: <type 'numpy.ndarray'>

旧解决方案

frequencyVectors.map(lambda vector: DenseVector(vector.toArray()))

编辑 1 - 代码可重现

from pyspark import SparkConf, SparkContext
from pyspark.sql import SQLContext, Row
from pyspark.sql.functions import split

from pyspark.ml.feature import CountVectorizer
from pyspark.mllib.clustering import LDA, LDAModel
from pyspark.mllib.linalg import Vectors
from pyspark.ml.feature import HashingTF, IDF, Tokenizer
from pyspark.mllib.linalg import SparseVector, DenseVector

sqlContext = SQLContext(sparkContext=spark.sparkContext, sparkSession=spark)
sc.setLogLevel('ERROR')

sentenceData = spark.createDataFrame([
    (0, "Hi I heard about Spark"),
    (0, "I wish Java could use case classes"),
    (1, "Logistic regression models are neat")
], ["label", "sentence"])
sentenceData = sentenceData.withColumn("sentence", split("sentence", "\s+"))
sentenceData.show()

vectorizer = CountVectorizer(inputCol="sentence", outputCol="rawfeatures").fit(sentenceData)
countVectors = vectorizer.transform(sentenceData).select("label", "rawfeatures")

idf = IDF(inputCol="rawfeatures", outputCol="features")
idfModel = idf.fit(countVectors)
tfidf = idfModel.transform(countVectors).select("label", "features")
frequencyDenseVectors = tfidf.rdd.map(lambda vector: [vector[0],DenseVector(vector[1].toArray())])
frequencyDenseVectors.map(lambda x: (x, )).toDF(["rawfeatures"])

Answer 1

您不能直接转换RDD[Vector]。它应该映射到可以解释为structs 的对象的RDD，例如RDD[Tuple[Vector]]：

frequencyDenseVectors.map(lambda x: (x, )).toDF(["rawfeatures"])

否则 Spark 将尝试转换对象 __dict__ 并创建使用不受支持的 NumPy 数组作为字段。

from pyspark.ml.linalg import DenseVector  
from pyspark.sql.types import _infer_schema

v = DenseVector([1, 2, 3])
_infer_schema(v)

TypeError                                 Traceback (most recent call last)
... 
TypeError: not supported type: <class 'numpy.ndarray'>

对比

_infer_schema((v, ))

StructType(List(StructField(_1,VectorUDT,true)))

注意事项：

• 在 Spark 2.0 中，您必须使用正确的本地类型：

  • pyspark.ml.linalg 工作时 DataFrame 基于 pyspark.ml API。
  • pyspark.mllib.linalg 工作时 RDD 基于 pyspark.mllib API。

  这两个命名空间不再兼容，需要显式转换（例如How to convert from org.apache.spark.mllib.linalg.VectorUDT to ml.linalg.VectorUDT）。

• 编辑中提供的代码与原始问题中的代码不同。您应该知道tuple 和list 的语义不同。如果你将向量映射成对使用tuple 并直接转换为DataFrame：

  tfidf.rdd.map(
      lambda row: (row[0], DenseVector(row[1].toArray()))
  ).toDF()
  

  使用tuple（产品类型）也适用于嵌套结构，但我怀疑这是您想要的：

  (tfidf.rdd
      .map(lambda row: (row[0], DenseVector(row[1].toArray())))
      .map(lambda x: (x, ))
      .toDF())
  

  list 在顶级row 之外的任何其他位置都被解释为ArrayType。

• 使用 UDF 进行转换更简洁 (Spark Python: Standard scaler error "Do not support ... SparseVector")。

Answer 2

我认为这里的问题是 createDataframe 不以denseVactor 作为参数请尝试将denseVector 转换为相应的集合[即数组或列表]。在 scala 和 java 中

toArray()

方法可用，您可以将denseVector转换为数组或列表，然后尝试创建dataFrame。