2

我有一个行矩阵xw

scala> xw
res109: org.apache.spark.mllib.linalg.distributed.RowMatrix = org.apache.spark.mllib.linalg.distributed.RowMatrix@8e74950

我想对其每个元素应用一个函数：

f(x)=exp(-x*x)

矩阵的元素类型可以可视化为：

scala> xw.rows.first

res110: org.apache.spark.mllib.linalg.Vector = [0.008930720313311474,0.017169380001300985,-0.013414238595719104,0.02239106636801034,0.023009502628798143,0.02891937604244297,0.03378470969100948,0.03644030110678057,0.0031586143217048825,0.011230244437457062,0.00477455053405408,0.020251682490519785,-0.005429788421130285,0.011578489275815267,0.0019301805575977788,0.022513736483645713,0.009475039307158668,0.019457912132044935,0.019209006632742498,-0.029811133879879596]

我的主要问题是我不能在矢量上使用地图

scala> xw.rows.map(row => row.map(e => breeze.numerics.exp(e)))
<console>:44: error: value map is not a member of org.apache.spark.mllib.linalg.Vector
              xw.rows.map(row => row.map(e => breeze.numerics.exp(e)))
                                     ^

scala>

我该如何解决？

4

1 回答 1

6

这是假设你知道你实际上有一个DenseVector（似乎是这样）。您可以调用toArray具有地图的矢量，然后将其转换回DenseVectorwith Vectors.dense：

xw.rows.map{row => Vectors.dense(row.toArray.map{e => breeze.numerics.exp(e)})}

你也可以这样做SparseVector；它在数学上是正确的，但转换为数组可能效率极低。另一种选择是调用row.copy然后使用foreachActive，这对密集和稀疏向量都有意义。但copy可能无法针对Vector您正在使用的特定类实现，而且如果您不知道向量的类型，则无法更改数据。如果你真的需要支持稀疏和密集的向量，我会这样做：

xw.rows.map{
  case denseVec: DenseVector => 
    Vectors.dense(denseVec.toArray.map{e => breeze.numerics.exp(e)})}
  case sparseVec: SparseVector =>
    //we only need to update values of the sparse vector -- the indices remain the same  
    val newValues: Array[Double] = sparseVec.values.map{e => breeze.numerics.exp(e)}
    Vectors.sparse(sparseVec.size, sparseVec.indices, newValues)
}
于 2015-02-12T15:26:17.890 回答