What is the MathNet equivalent of numpy array[0,:] *= 1.23

Question

I have to port some python (numpy) code in c# (MathNet). I can write in python:

mtx = np.array([[0,1,2],[3,4,5]])
mtx[0,:] *= 1.23    #multiply all elements in row 0 by 1.23

How can I do this in MathNet? Is there better (faster) solution than:

 Matrix<double> mtx = Matrix<double>.Build.Dense(2,3);
 //...
 for(int i = 0; i < mtx.ColumnCount; i++)
    mtx[0,i] *= 1.23;

?

score 5 · Answer 1 · answered May 22 '16 at 10:50

5

For completeness: Math.NET Numerics itself actually does support a notation which is somewhat close to your NumPy example. C# does not support it, but other more powerful .Net languages like F# do:

let mtx = matrix [[0.;1.;2.];[3.;4.;5.]]
mtx.[0,*] <- 1.23 * mtx.[0,*]

answered May 22 '16 at 10:50

Christoph Rüegg

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Do you think, if I have to use the library often and more intensively , I should prefer the F# language? – ThWin May 22 '16 at 12:28
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@ThWin, I think, yes. – FoggyFinder May 22 '16 at 20:26

score 2 · Accepted Answer · edited Jun 20 '20 at 09:12

2

There are few ways, for sure cleaner than for. Starting with matrix full of 1.

 Matrix<double> mtx = Matrix<double>.Build.Dense(2, 3, 1);
 
 mtx.SetRow(0, mtx.Row(0).Multiply(1.23));

 Console.WriteLine(mtx);

returns

DenseMatrix 2x3-Double

1,23 1,23 1,23

1 1 1

edited Jun 20 '20 at 09:12

Community

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answered May 20 '16 at 21:17

Piotr Leniartek

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What is the MathNet equivalent of numpy array[0,:] *= 1.23

2 Answers2