Note that also you can use boost.ublas for matrix creation and manipulation and also boost.graph to represent and manipulate graphs in a number of ways, as well as using algorithms on them, etc. Edit: Anyway, doing a range-check version of a vector for your purposes is not a hard thing: template <typename T> class BoundsMatrix … Read more
The ndgrid function almost gives the answer, but has one caveat: n output variables must be explicitly defined to call it. Since n is arbitrary, the best way is to use a comma-separated list (generated from a cell array with ncells) to serve as output. The resulting n matrices are then concatenated into the desired … Read more
You can also check if all the eigenvalues of matrix are positive. If so, the matrix is positive definite: import numpy as np def is_pos_def(x): return np.all(np.linalg.eigvals(x) > 0)
SVD can be understood from a geometric sense for square matrices as a transformation on a vector. Consider a square n x n matrix M multiplying a vector v to produce an output vector w: w = M*v The singular value decomposition M is the product of three matrices M=U*S*V, so w=U*S*V*v. U and V … Read more
No. A double** is a pointer to a pointer to a double (double*). So actually it should be created like this (note the extra * in the first malloc sizeof()): double** result = (double**) malloc(sizeof(double*)*4); for (int i = 0; i < 4; i++) { result[i] = (double*) malloc(sizeof(double)*4); } So in memory it would … Read more
There is a new numpy function in version 1.7.0 numpy.pad that can do this in one-line. Like the other answers, you can construct the diagonal matrix with np.diag before the padding. The tuple ((0,N),(0,0)) used in this answer indicates the “side” of the matrix which to pad. import numpy as np A = np.array([1, 2, … Read more
nota bene. This answer has 20 upvotes now, but it is not intended as an endorsement of std::valarray. In my experience, time is better spent installing and learning to use a full-fledged math library such as Eigen. Valarray has fewer features than the competition, but it isn’t more efficient or particularly easier to use. If … Read more
Consider the code: M = randi(100, [3 4]); %# input matrix ind = reshape(1:numel(M), size(M)); %# indices of elements ind = fliplr( spdiags( fliplr(ind) ) ); %# get the anti-diagonals ind(:,1:2:end) = flipud( ind(:,1:2:end) ); %# reverse order of odd columns ind(ind==0) = []; %# keep non-zero indices M(ind) %# get elements in zigzag order … Read more