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syntax for armadillo into

Feb 17, 2015 at 8:19 PM
I am a new to Math.Net Numerics and am struck in using the following in Math.NEt

from armadillo c++ program
rowvec td;
colvec dda = ( - td(0)) * 5;
dda = dda .rows(1, 4- 1);

Using Math.Net in c#
Matrix<double> td;
Matrix<double> dda = (td.Transpose() - td[0,0] * 5);
dda = dda .rows(1, 4- 1); //this line I am not able to convert may be like dda = dda[1,3] doesn't work as it returns a double

What is the best way to convert ...
Feb 17, 2015 at 9:07 PM
I don't know Armadillo well, but it sounds more like:
Vector<double> td;
Vector<double> dda = ((td - td[0]) * 5);
dda = dda.SubVector(1,3);