Question

我已经在Scala中初始化了两个矩阵（X和Y），如下所示，

 var x = ofDim[Int](a1,b1)
 var y = ofDim[Int](a2,b2)

x，y，a1，a2，b1和b2是变量。现在我需要通过Y（X / Y）来决定X.怎么能实现呢？

Answer 1

还有其他使用Apache Commons的方法。但是，重要的是观察除法运算应用乘法运算和反演运算，并且某些矩阵是可逆的，而其他矩阵则是：https://en.wikipedia.org/wiki/Invertible_matrix

以下示例应用Apache Commons库（ Study.scala ）：

import org.apache.commons.math3.linear._

object Study {

  def main(args: Array[String]): Unit = {

     val xArray = Array(Array(1.0, 2.0), Array(3.0, 4.0))
     val yArray = Array(Array(1.0, 2.0), Array(3.0, 4.0))

     val x = new Array2DRowRealMatrix(xArray)
     val y = new Array2DRowRealMatrix(yArray)

     val yInverse = new LUDecomposition(y).getSolver().getInverse();
     val w = x.multiply(yInverse)

     for(i <- 0 until w.getRowDimension())
         for(j <- 0 until w.getColumnDimension())
             println(w.getEntry(i, j))

    }
}

提示：如果您打算使用scala控制台，则需要指定classpath ...

 scala -classpath .../commons-math3/3.2/commons-math3-3.2.jar

...在scala会话中加载算法......

  :load .../Study.scala

...结果出来调用Study的主要功能（可以应用近似值）......

 scala> Study.main(null)

0.99 / 1.11E-16 / 0.0 / 1.02

Answer 2

尝试：

let imageCache = NSCache<NSString, UIImage>()

extension UIImageView {

func loadImageUsingCacheWithUrlString(_ urlString: String) {

    self.image = nil

    //check cache for image first
    if let cachedImage = imageCache.object(forKey: urlString as NSString){
        self.image = cachedImage
        return
    }

    //otherwise fire off a new download
    guard let url = URL(string: urlString) else {
        return
    }
    URLSession.shared.dataTask(with: url) {
        data, response, error in

        //download hit an error so lets return out
        if error != nil {

            return
        }
        DispatchQueue.main.async(execute: {
            if let downloadedImage = UIImage(data: data!) {
                imageCache.setObject(downloadedImage, forKey: urlString as NSString)
                self.image = downloadedImage
            }
        })
        }.resume()
   }

}

Answer 3

图书馆Beeze，正如其他回应中所说，是必要的。您可以使用SBT或Maven
安装它

Breeze项目可以从GitHub

下载

这是Maven方法：

<dependency>
    <groupId>org.scalanlp</groupId>
    <artifactId>breeze_2.10</artifactId> <!-- or 2.11 -->
    <version>0.12</version>
</dependency>

代码......

import breeze.linalg.DenseMatrix

object Division {

    def main(args: Array[String]): Unit = {

        var a1 = 10
        var a2 = 11
        var b1 = 12
        var b2 = 13

       //var x = Array.ofDim[Int](a1,b1)
       //var y = Array.ofDim[Int](a2,b2)

       var x = DenseMatrix(a1,b1)
       var y = DenseMatrix(a2,b2)

       var result = x/y
       print(result)
  }
}

如何在Scala中用另一个矩阵划分Matrix

3 个答案: