稀疏向量 pyspark

Question

我想找到一种使用数据帧在 PySpark 中创建备用向量的有效方法。

假设给定事务输入：

df = spark.createDataFrame([
    (0, "a"),
    (1, "a"),
    (1, "b"),
    (1, "c"),
    (2, "a"),
    (2, "b"),
    (2, "b"),
    (2, "b"),
    (2, "c"),
    (0, "a"),
    (1, "b"),
    (1, "b"),
    (2, "cc"),
    (3, "a"),
    (4, "a"),
    (5, "c")
], ["id", "category"])

+---+--------+
| id|category|
+---+--------+
|  0|       a|
|  1|       a|
|  1|       b|
|  1|       c|
|  2|       a|
|  2|       b|
|  2|       b|
|  2|       b|
|  2|       c|
|  0|       a|
|  1|       b|
|  1|       b|
|  2|      cc|
|  3|       a|
|  4|       a|
|  5|       c|
+---+--------+

总结格式：

df.groupBy(df["id"],df["category"]).count().show()

+---+--------+-----+
| id|category|count|
+---+--------+-----+
|  1|       b|    3|
|  1|       a|    1|
|  1|       c|    1|
|  2|      cc|    1|
|  2|       c|    1|
|  2|       a|    1|
|  1|       a|    1|
|  0|       a|    2|
+---+--------+-----+

我的目标是通过 id 得到这个输出：

+---+-----------------------------------------------+
| id|                                       feature |
+---+-----------------------------------------------+
|  2|SparseVector({a: 1.0, b: 3.0, c: 1.0, cc: 1.0})|

你能指出我正确的方向吗？使用 Java 中的 mapreduce 对我来说似乎更容易。

Answer 1

这可以通过pivot 和VectorAssembler 轻松完成。将聚合替换为pivot：

 pivoted = df.groupBy("id").pivot("category").count().na.fill(0)

组装：

from pyspark.ml.feature import VectorAssembler

input_cols = [x for x in pivoted.columns if x != id]

result = (VectorAssembler(inputCols=input_cols, outputCol="features")
    .transform(pivoted)
    .select("id", "features"))

结果如下。这将根据稀疏性选择更有效的表示：

+---+---------------------+
|id |features             |
+---+---------------------+
|0  |(5,[1],[2.0])        |
|5  |(5,[0,3],[5.0,1.0])  |
|1  |[1.0,1.0,3.0,1.0,0.0]|
|3  |(5,[0,1],[3.0,1.0])  |
|2  |[2.0,1.0,3.0,1.0,1.0]|
|4  |(5,[0,1],[4.0,1.0])  |
+---+---------------------+

当然您仍然可以将其转换为单个表示：

from pyspark.ml.linalg import SparseVector, VectorUDT
import numpy as np

def to_sparse(c):
    def to_sparse_(v):
        if isinstance(v, SparseVector):
            return v
        vs = v.toArray()
        nonzero = np.nonzero(vs)[0]
        return SparseVector(v.size, nonzero, vs[nonzero])
    return udf(to_sparse_, VectorUDT())(c)

+---+-------------------------------------+
|id |features                             |
+---+-------------------------------------+
|0  |(5,[1],[2.0])                        |
|5  |(5,[0,3],[5.0,1.0])                  |
|1  |(5,[0,1,2,3],[1.0,1.0,3.0,1.0])      |
|3  |(5,[0,1],[3.0,1.0])                  |
|2  |(5,[0,1,2,3,4],[2.0,1.0,3.0,1.0,1.0])|
|4  |(5,[0,1],[4.0,1.0])                  |
+---+-------------------------------------+

Answer 2

如果您将数据框转换为 RDD，则可以遵循类似 mapreduce 的框架 reduceByKey。这里唯一真正棘手的部分是格式化 spark 的日期 sparseVector

导入包，创建数据

from pyspark.ml.feature import StringIndexer
from pyspark.ml.linalg import Vectors
df = sqlContext.createDataFrame([
    (0, "a"),
    (1, "a"),
    (1, "b"),
    (1, "c"),
    (2, "a"),
    (2, "b"),
    (2, "b"),
    (2, "b"),
    (2, "c"),
    (0, "a"),
    (1, "b"),
    (1, "b"),
    (2, "cc"),
    (3, "a"),
    (4, "a"),
    (5, "c")
], ["id", "category"])

为类别创建数字表示（稀疏向量需要）

indexer = StringIndexer(inputCol="category", outputCol="categoryIndex")
df = indexer.fit(df).transform(df) 

按索引分组，获取计数

df = df.groupBy(df["id"],df["categoryIndex"]).count()

转换为rdd，将数据映射到id & [categoryIndex, count]的键值对

rdd = df.rdd.map(lambda x: (x.id, [(x.categoryIndex, x['count'])]))

通过 key 减少以获得 id 的键值对和该 id 的所有 [categoryIndex, count] 的列表

rdd = rdd.reduceByKey(lambda a, b: a + b)

映射数据，将每个id的所有[categoryIndex, count]列表转化为稀疏向量

rdd = rdd.map(lambda x: (x[0], Vectors.sparse(len(x[1]), x[1])))

转换回数据框

finalDf = sqlContext.createDataFrame(rdd, ['id', 'feature'])

数据检查

finalDf.take(5)

 [Row(id=0, feature=SparseVector(1, {1: 2.0})),
  Row(id=1, feature=SparseVector(3, {0: 3.0, 1: 1.0, 2: 1.0})),
  Row(id=2, feature=SparseVector(4, {0: 3.0, 1: 1.0, 2: 1.0, 3: 1.0})),
  Row(id=3, feature=SparseVector(1, {1: 1.0})),
  Row(id=4, feature=SparseVector(1, {1: 1.0}))]