2015-02-10 119 views
2

我有一個rowMatrix xw火花mllib應用功能

scala> xw 
res109: org.apache.spark.mllib.linalg.distributed.RowMatrix = [email protected] 

,我想給一個函數應用到它的每個元素的一個rowMatrix的所有元素:

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

的矩陣元素的類型可以被可視化爲:

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] 

我的主要問題是我不能在地圖上使用地圖

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> 

我該如何解決?

回答

6

這是假設你知道你實際上有一個DenseVector(這似乎是這種情況)。您可以在載體中,其中有一個叫圖toArray,然後再轉換回DenseVectorVectors.dense

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

你可以這樣做一個SparseVector爲好;它在數學上是正確的,但是轉換爲數組可能效率極低。另一個選擇是撥打row.copy,然後使用foreachActive,這對密集和稀疏矢量都有意義。但copy可能不會針對您正在使用的特定Vector類實現,並且如果您不知道向量的類型,則不能對數據進行變異。如果你真的需要支持稀疏密集的向量,我會做這樣的事情:

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) 
} 
+0

感謝您的答案。所以對於vectors.dense類,你建議我使用提供的代碼行嗎?你是否可以在答案的第二部分編寫代碼?我是斯卡拉初學者,所以它不是太容易遵循 – Donbeo 2015-02-12 15:36:16

+0

@唐貝我更新了答案一點。如果你確定你有DenseVectors,那就去找第一個答案。如果你可能稀疏或密集,你可以使用第二個,但請注意,即使這樣也不能處理Vector的其他可能的實現。 (例如,它不處理'VectorUDT') – 2015-02-12 17:58:13