如何将向量拆分为列-使用PySp

Question

上下文：我有一个包含两列的DataFrame：word和vector。其中“vector”的列类型是VectorUDT。

例如：

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

我想得到这个：

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

问题：

如何使用PySpark为每个维度将一列向量分成若干列？

提前谢谢

Answer 1

一种可能的方法是转换到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自定义项：

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|
## +-------+-----+-----+-----+

关于Scala等价物，请参见Spark Scala: How to convert Dataframe[vector] to DataFrame[f1:Double, ..., fn: Double)]。

Answer 2

def splitVecotr(df, new_features=['f1','f2']):
schema = df.schema
cols = df.columns

for col in new_features: # new_features should be the same length as vector column length
    schema = schema.add(col,DoubleType(),True)

return spark.createDataFrame(df.rdd.map(lambda row: [row[i] for i in cols]+row.features.tolist()), schema)

函数将特征向量列转换为单独的列

Answer 3

使用how-to-access-element-of-a-vectorudt-column-in-a-spark-dataframe中的自定义项要快得多

上面zero323在解决方案中给出的extract函数使用to list，它创建一个Python列表对象，用Python float对象填充它，通过遍历列表来找到所需的元素，然后需要将列表转换回java double；对每一行重复。使用rdd要比tou数组udf慢得多，后者也调用toList，但两者都比让SparkSQL处理大部分工作的udf慢得多。

比较rdd extract和此处提议的数组udf与来自3955864的udf的计时代码：

from pyspark.context import SparkContext
from pyspark.sql import Row, SQLContext, SparkSession
from pyspark.sql.functions import lit, udf, col
from pyspark.sql.types import ArrayType, DoubleType
import pyspark.sql.dataframe
from pyspark.sql.functions import pandas_udf, PandasUDFType

sc = SparkContext('local[4]', 'FlatTestTime')

spark = SparkSession(sc)
spark.conf.set("spark.sql.execution.arrow.enabled", True)

from pyspark.ml.linalg import Vectors

# copy the two rows in the test dataframe a bunch of times,
# make this small enough for testing, or go for "big data" and be prepared to wait
REPS = 20000

df = sc.parallelize([
    ("assert", Vectors.dense([1, 2, 3]), 1, Vectors.dense([4.1, 5.1])),
    ("require", Vectors.sparse(3, {1: 2}), 2, Vectors.dense([6.2, 7.2])),
] * REPS).toDF(["word", "vector", "more", "vorpal"])

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

def test_extract():
    return df.rdd.map(extract).toDF(['word', 'vector__0', 'vector__1', 'vector__2', 'more', 'vorpal__0', 'vorpal__1'])

def to_array(col):
    def to_array_(v):
        return v.toArray().tolist()
    return udf(to_array_, ArrayType(DoubleType()))(col)

def test_to_array():
    df_to_array = df.withColumn("xs", to_array(col("vector"))) \
        .select(["word"] + [col("xs")[i] for i in range(3)] + ["more", "vorpal"]) \
        .withColumn("xx", to_array(col("vorpal"))) \
        .select(["word"] + ["xs[{}]".format(i) for i in range(3)] + ["more"] + [col("xx")[i] for i in range(2)])
    return df_to_array

# pack up to_array into a tidy function
def flatten(df, vector, vlen):
    fieldNames = df.schema.fieldNames()
    if vector in fieldNames:
        names = []
        for fieldname in fieldNames:
            if fieldname == vector:
                names.extend([col(vector)[i] for i in range(vlen)])
            else:
                names.append(col(fieldname))
        return df.withColumn(vector, to_array(col(vector)))\
                 .select(names)
    else:
        return df

def test_flatten():
    dflat = flatten(df, "vector", 3)
    dflat2 = flatten(dflat, "vorpal", 2)
    return dflat2

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

ith = udf(ith_, DoubleType())

select = ["word"]
select.extend([ith("vector", lit(i)) for i in range(3)])
select.append("more")
select.extend([ith("vorpal", lit(i)) for i in range(2)])

# %% timeit ...
def test_ith():
    return df.select(select)

if __name__ == '__main__':
    import timeit

    # make sure these work as intended
    test_ith().show(4)
    test_flatten().show(4)
    test_to_array().show(4)
    test_extract().show(4)

    print("i_th\t\t",
          timeit.timeit("test_ith()",
                       setup="from __main__ import test_ith",
                       number=7)
         )
    print("flatten\t\t",
          timeit.timeit("test_flatten()",
                       setup="from __main__ import test_flatten",
                       number=7)
         )
    print("to_array\t",
          timeit.timeit("test_to_array()",
                       setup="from __main__ import test_to_array",
                       number=7)
         )
    print("extract\t\t",
          timeit.timeit("test_extract()",
                       setup="from __main__ import test_extract",
                       number=7)
         )

结果：

i_th         0.05964796099999958
flatten      0.4842299350000001
to_array     0.42978780299999997
extract      2.9254476840000017