如何将 Vector 拆分为列 - 使用 PySpark [重复]

问题描述：

上下文：我有一个DataFrame包含 2 列的数据：单词和向量。其中“向量”的列类型是VectorUDT。

一个例子：

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

我想要得到这个：

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

问题：

如何使用 PySpark 将一列向量拆分为每一维的几列？

提前致谢

解决方案 1：

Spark >= 3.0.0

从 Spark 3.0.0 开始，无需使用 UDF 即可完成此操作。

from pyspark.ml.functions import vector_to_array

(df
    .withColumn("xs", vector_to_array("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|
## +-------+-----+-----+-----+

Spark < 3.0.0

一种可能的方法是转换为 RDD 或从 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一个 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|
## +-------+-----+-----+-----+

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

解决方案 2：

rawPrediction要将训练 PySpark ML 模型后生成的或列拆分probability为 Pandas 列，可以像这样拆分：

your_pandas_df['probability'].apply(lambda x: pd.Series(x.toArray()))

解决方案 3：

使用how-to-access-element-of-a-vectorudt-column-in-a-spark-dataframe 中的第 i 个 udf 要快得多

上述 zero323 的解决方案中提供的 extract 函数使用 toList，它创建一个 Python 列表对象，用 Python 浮点对象填充它，通过遍历列表找到所需的元素，然后需要将其转换回 java double；对每一行重复此操作。使用 rdd 比 to_array udf（也调用 toList）慢得多，但两者都比让 SparkSQL 处理大部分工作的 udf 慢得多。

将此处提出的 rdd extract 和 to_array udf 与3955864中的 i_th udf 进行比较的时间代码：

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"))) \n        .select(["word"] + [col("xs")[i] for i in range(3)] + ["more", "vorpal"]) \n        .withColumn("xx", to_array(col("vorpal"))) \n        .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)))\n                 .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        ",
          timeit.timeit("test_ith()",
                       setup="from __main__ import test_ith",
                       number=7)
         )
    print("flatten        ",
          timeit.timeit("test_flatten()",
                       setup="from __main__ import test_flatten",
                       number=7)
         )
    print("to_array    ",
          timeit.timeit("test_to_array()",
                       setup="from __main__ import test_to_array",
                       number=7)
         )
    print("extract        ",
          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

解决方案 4：

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)

该函数将特征向量列转换为单独的列