How to access element of a VectorUDT column in a Spark DataFrame?

Question

I have a dataframe df with a VectorUDT column named features. How do I get an element of the column, say first element?

I've tried doing the following

from pyspark.sql.functions import udf
first_elem_udf = udf(lambda row: row.values[0])
df.select(first_elem_udf(df.features)).show()

but I get a net.razorvine.pickle.PickleException: expected zero arguments for construction of ClassDict(for numpy.dtype) error. Same error if I do first_elem_udf = first_elem_udf(lambda row: row.toArray()[0]) instead.

I also tried explode() but I get an error because it requires an array or map type.

This should be a common operation, I think.

Answer 1

Convert output to float:

from pyspark.sql.types import DoubleType
from pyspark.sql.functions import lit, udf

def ith_(v, i):
    try:
        return float(v[i])
    except ValueError:
        return None

ith = udf(ith_, DoubleType())

Example usage:

from pyspark.ml.linalg import Vectors

df = sc.parallelize([
    (1, Vectors.dense([1, 2, 3])),
    (2, Vectors.sparse(3, [1], [9]))
]).toDF(["id", "features"])

df.select(ith("features", lit(1))).show()

## +-----------------+
## |ith_(features, 1)|
## +-----------------+
## |              2.0|
## |              9.0|
## +-----------------+

Explanation:

Output values have to be reserialized to equivalent Java objects. If you want to access values (beware of SparseVectors) you should use item method:

v.values.item(0)

which return standard Python scalars. Similarly if you want to access all values as a dense structure:

v.toArray().tolist()