如何将Vector拆分为列 - 使用PySpark

Question

上下文:我有DataFrame2列:单词和向量.其中"vector"的列类型是VectorUDT.

一个例子:

word    |  vector
assert  | [435,323,324,212...]

我希望得到这个:

word   |  v1 | v2  | v3 | v4 | v5 | v6 ......
assert | 435 | 5435| 698| 356|....

题:

如何使用PySpark为每个维度拆分包含多列向量的列？

提前致谢

Answer 1

一种可能的方法是转换为RDD和从RDD转换:

from pyspark.ml.linalg import Vectors

df = sc.parallelize([
    ("assert", Vectors.dense([1, 2, 3])),
    ("require", Vectors.sparse(3, {1: 2}))
]).toDF(["word", "vector"])

def extract(row):
    return (row.word, ) + tuple(row.vector.toArray().tolist())

df.rdd.map(extract).toDF(["word"])  # Vector values will be named _2, _3, ...

## +-------+---+---+---+
## |   word| _2| _3| _4|
## +-------+---+---+---+
## | assert|1.0|2.0|3.0|
## |require|0.0|2.0|0.0|
## +-------+---+---+---+

另一种解决方案是创建UDF:

from pyspark.sql.functions import udf, col
from pyspark.sql.types import ArrayType, DoubleType

def to_array(col):
    def to_array_(v):
        return v.toArray().tolist()
    # Important: asNondeterministic requires Spark 2.3 or later
    # It can be safely removed i.e.
    # return udf(to_array_, ArrayType(DoubleType()))(col)
    # but at the cost of decreased performance
    return udf(to_array_, ArrayType(DoubleType())).asNondeterministic()(col)

(df
    .withColumn("xs", to_array(col("vector")))
    .select(["word"] + [col("xs")[i] for i in range(3)]))

## +-------+-----+-----+-----+
## |   word|xs[0]|xs[1]|xs[2]|
## +-------+-----+-----+-----+
## | assert|  1.0|  2.0|  3.0|
## |require|  0.0|  2.0|  0.0|
## +-------+-----+-----+-----+

对于Scala等效,请参阅Spark Scala:如何将Dataframe [vector]转换为DataFrame [f1:Double,...,fn:Double]].