PySpark - SparseVector Column to Matrix

Question

I am very new to using PySpark. I have a column of SparseVectors in my PySpark dataframe.

rescaledData.select('features').show(5,False)

+--------------------------------------------------------------------------------------------------------------------------------------------------------------------+
|features                                                                                                                                                            |
+--------------------------------------------------------------------------------------------------------------------------------------------------------------------+
|(262144,[43953,62425,66522,148962,174441,249180],[3.9219733362813143,3.9219733362813143,1.213923135179104,3.9219733362813143,3.9219733362813143,0.5720692490067093])|
|(262144,[57925,66522,90939,249180],[3.5165082281731497,1.213923135179104,3.9219733362813143,0.5720692490067093])                                                    |
|(262144,[23366,45531,73408,211290],[2.6692103677859462,3.005682604407159,3.5165082281731497,3.228826155721369])                                                     |
|(262144,[30913,81939,99546,137643,162885,249180],[3.228826155721369,3.9219733362813143,3.005682604407159,3.005682604407159,3.228826155721369,1.1441384980134186])   |
|(262144,[108134,152329,249180],[3.9219733362813143,2.6692103677859462,2.8603462450335466])                                                                          |
+--------------------------------------------------------------------------------------------------------------------------------------------------------------------+

I need to convert the above dataframe into a matrix where every row in the matrix corresponds to a SparseVector in that exact row in the dataframe.

for example,

+-----------------+
|features         |
+-----------------+
|(7,[1,2],[45,63])|
|(7,[3,5],[85,69])|
|(7,[1,2],[89,56])|
+-----------------+

Must be converted to

[[0,45,63,0,0,0,0]
[0,0,0,85,0,69,0]
[0,89,56,0,0,0,0]]

I have read the link below, which shows that there is a function toArray() which does exactly what I want. https://mingchen0919.github.io/learning-apache-spark/pyspark-vectors.html

However, I am having trouble using it.

vector_udf = udf(lambda vector: vector.toArray())
rescaledData.withColumn('features_', vector_udf(rescaledData.features)).first()

I need it to convert every row into an array and then convert the PySpark dataframe into a matrix.

None of the answers good enough to accept, or at least upvote as useful?? — desertnaut
– desertnaut, Commented Dec 8, 2017 at 10:38

Suresh · Accepted Answer · 2017-12-06 14:56:19Z

7

toArray() will return numpy array. we can convert to list and then collect the dataframe.

from pyspark.sql.types import *
vector_udf = udf(lambda vector: vector.toArray().tolist(),ArrayType(DoubleType()))

df.show() ## my sample dataframe
+-------------------+
|           features|
+-------------------+
|(4,[1,3],[3.0,4.0])|
|(4,[1,3],[3.0,4.0])|
|(4,[1,3],[3.0,4.0])|
+-------------------+

colvalues = df.select(vector_udf('features').alias('features')).collect()

list(map(lambda x:x.features,colvalues))
[[0.0, 3.0, 0.0, 4.0], [0.0, 3.0, 0.0, 4.0], [0.0, 3.0, 0.0, 4.0]]

answered Dec 6, 2017 at 14:56

Suresh

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Alper t. Turker · Accepted Answer · 2017-12-06 13:19:12Z

7

Convert to RDD and map:

vectors = df.select("features").rdd.map(lambda row: row.features)

Convert result to distributed matrix:

from pyspark.mllib.linalg.distributed import RowMatrix

matrix = RowMatrix(vectors)

If you want DenseVectors (memory requirements!):

vectors = df.select("features").rdd.map(lambda row: row.features.toArray())

answered Dec 6, 2017 at 13:19

Alper t. Turker

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