PySpark 2.2.0 : 'numpy.ndarray' object has no attribute 'indices'

Question

Task

I'm calculating the size on the indices within a __SparseVector__ using Python API for Spark (PySpark).

Script

def score_clustering(dataframe):
assembler = VectorAssembler(inputCols = dataframe.drop("documento").columns, outputCol = "variables")
data_transformed = assembler.transform(dataframe)
data_transformed_rdd = data_transformed.select("documento", "variables").orderBy(data_transformed.documento.asc()).rdd
count_variables = data_transformed_rdd.map(lambda row : [row[0], row[1].indices.size]).toDF(["id", "frequency"])

Issue

When I execute the action __.count()__ on the __count_variables__ dataframe an error shows up:

AttributeError: 'numpy.ndarray' object has no attribute 'indices'

The main part to consider is:

data_transformed_rdd.map(lambda row : [row[0], row[1].indices.size]).toDF(["id", "frequency"])

I believe this chunk has to do with the error, but I cannot understand why the exception is telling about __numpy.ndarray__ if I'm doing the calculations through mapping that __lambda expression__ whose taking as argument a __SparseVector__ (created with the __assembler__).

Any suggestions? Does anyone maybe know what I'm doing wrong?

Answer 1

There are two problems here. The first one is in indices.size call, indices and size are two different attributes of SparseVector class, size is the complete vector size and indices are the vector indices whose values are non-zero, but size is not a indices attribute. So, assuming that all your vectors are instances of SparseVector class:

from pyspark.ml.linalg import Vectors

df = spark.createDataFrame([(0, Vectors.sparse(4, [0, 1], [11.0, 2.0])),
                            (1, Vectors.sparse(4, [], [])),
                            (3, Vectors.sparse(4, [0,1,2], [2.0, 2.0, 2.0]))],
                           ["documento", "variables"])

df.show()

+---------+--------------------+
|documento|           variables|
+---------+--------------------+
|        0|(4,[0,1],[11.0,2.0])|
|        1|           (4,[],[])|
|        3|(4,[0,1,2],[2.0,2...|
+---------+--------------------+

The solution is len function:

df = df.rdd.map(lambda x: (x[0], x[1], len(x[1].indices)))\
               .toDF(["documento", "variables", "frecuencia"])
df.show()  
+---------+--------------------+----------+
|documento|           variables|frecuencia|
+---------+--------------------+----------+
|        0|(4,[0,1],[11.0,2.0])|         2|
|        1|           (4,[],[])|         0|
|        3|(4,[0,1,2],[2.0,2...|         3|
+---------+--------------------+----------+

And here comes the second problem: VectorAssembler does not always generate SparseVectors, depending on what is more efficient, SparseVector or DenseVectors can be generated (based on the number of zeros that your original vector has). For example, suppose the next data frame:

df = spark.createDataFrame([(0, Vectors.sparse(4, [0, 1], [11.0, 2.0])),
                             (1, Vectors.dense([1., 1., 1., 1.])),
                              (3, Vectors.sparse(4, [0,1,2], [2.0, 2.0, 2.0]))], 
                           ["documento", "variables"])

df.show()      
+---------+--------------------+
|documento|           variables|
+---------+--------------------+
|        0|(4,[0,1],[11.0,2.0])|
|        1|   [1.0,1.0,1.0,1.0]|
|        3|(4,[0,1,2],[2.0,2...|
+---------+--------------------+

The document 1 is a DenseVector and the previos solution does not work because DenseVectors has not indices attribute, so you have to use a more general representation of vectors to work with a DataFrame which contains both sparse and dense vectors, for example numpy:

import numpy as np
df = df.rdd.map(lambda x: (x[0], 
                           x[1], 
                           np.nonzero(x[1])[0].size))\
                .toDF(["documento", "variables", "frecuencia"])
df.show() 
+---------+--------------------+----------+
|documento|           variables|frecuencia|
+---------+--------------------+----------+
|        0|(4,[0,1],[11.0,2.0])|         2|
|        1|   [1.0,1.0,1.0,1.0]|         4|
|        3|(4,[0,1,2],[2.0,2...|         3|
+---------+--------------------+----------+