Convert Sparse Vector to Dense Vector in Pyspark

Question

I have a sparse vector like this

>>> countVectors.rdd.map(lambda vector: vector[1]).collect()
[SparseVector(13, {0: 1.0, 2: 1.0, 3: 1.0, 6: 1.0, 8: 1.0, 9: 1.0, 10: 1.0, 12: 1.0}), SparseVector(13, {0: 1.0, 1: 1.0, 2: 1.0, 4: 1.0}), SparseVector(13, {0: 1.0, 1: 1.0, 3: 1.0, 4: 1.0, 7: 1.0}), SparseVector(13, {1: 1.0, 2: 1.0, 5: 1.0, 11: 1.0})]

I am trying to convert this into dense vector in pyspark 2.0.0 like this

>>> frequencyVectors = countVectors.rdd.map(lambda vector: vector[1])
>>> frequencyVectors.map(lambda vector: Vectors.dense(vector)).collect()

I am getting an error like this:

16/12/26 14:03:35 ERROR Executor: Exception in task 0.0 in stage 13.0 (TID 13)
org.apache.spark.api.python.PythonException: Traceback (most recent call last):
  File "/opt/BIG-DATA/spark-2.0.0-bin-hadoop2.7/python/lib/pyspark.zip/pyspark/worker.py", line 172, in main
    process()
  File "/opt/BIG-DATA/spark-2.0.0-bin-hadoop2.7/python/lib/pyspark.zip/pyspark/worker.py", line 167, in process
    serializer.dump_stream(func(split_index, iterator), outfile)
  File "/opt/BIG-DATA/spark-2.0.0-bin-hadoop2.7/python/lib/pyspark.zip/pyspark/serializers.py", line 263, in dump_stream
    vs = list(itertools.islice(iterator, batch))
  File "<stdin>", line 1, in <lambda>
  File "/opt/BIG-DATA/spark-2.0.0-bin-hadoop2.7/python/lib/pyspark.zip/pyspark/mllib/linalg/__init__.py", line 878, in dense
    return DenseVector(elements)
  File "/opt/BIG-DATA/spark-2.0.0-bin-hadoop2.7/python/lib/pyspark.zip/pyspark/mllib/linalg/__init__.py", line 286, in __init__
    ar = np.array(ar, dtype=np.float64)
  File "/opt/BIG-DATA/spark-2.0.0-bin-hadoop2.7/python/lib/pyspark.zip/pyspark/ml/linalg/__init__.py", line 701, in __getitem__
    raise ValueError("Index %d out of bounds." % index)
ValueError: Index 13 out of bounds.

How can I achieve this conversion? Is there anything wrong here?

Set

Answer 1

This resolved my issue

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

Set

Answer 2

# to convert spark vector column in pyspark dataframe to dense vector 
from pyspark.ml.linalg import DenseVector

@udf(T.ArrayType(T.FloatType()))

def  toDense(v):

    v = DenseVector(v)

    new_array = list([float(x) for x in v])

    return new_array 

df.withColumn('features',toDense('features')).show()
#here 'features' column is vector type

Borkowski

you need to edit this to fix your indentation – Araarab

Welcome to Stack Overflow. While this code may answer the question, providing additional context regarding why and/or how this code answers the question improves its long-term value. How to Answer – Russo

Answer 3

If your PySpark DataFrame is of DataFrame[SparseVector], the following is what works for me:

df2=df.select("features")
from pyspark.ml.feature import VectorAssembler
assembler = VectorAssembler(inputCols=feat_cols, outputCol="features_dense")
df3 = assembler.transform(df2).select('features_dense')

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