It looks like glm 0.9.6 supports matrix decomposition http://glm.g-truc.net/0.9.6/api/a00204.html #include <glm/gtx/matrix_decompose.hpp> glm::mat4 transformation; // your transformation matrix. glm::vec3 scale; glm::quat rotation; glm::vec3 translation; glm::vec3 skew; glm::vec4 perspective; glm::decompose(transformation, scale, rotation, translation, skew, perspective);
For the general case, where your search string can be in any column, you can do this: >>> rows, cols = np.where(t == ‘bar’) >>> t[rows] array([[‘2’, ‘3’, ‘4’, ‘bar’], [‘8’, ‘9’, ‘1’, ‘bar’]], dtype=”|S11″)
float (*somethingAsMatrix)[2] = (float (*)[2]) matrixReturnAsArray;
There are several differences between a cell array and a matrix in MATLAB: A cell array may contain any arbitrary type of element in each cell; while a matrix requires the types of its elements to be homogeneous i.e. of the same type. As far as memory layout goes, all elements of a matrix are … Read more
Short answer: cv::Mat uses the heap to store its data, while cv::Matx uses the stack. A cv::Mat uses dynamic memory allocation (on the heap). This is appropriate for big matrices (like images) and lets you do things like shallow copies of a matrix, which is the default behavior of cv::Mat. However, for the small matrices … Read more
From top left to bottom right var theArray = [“ABCD”,”EFGH”,”IJKL”]; var length = { “x” : theArray[0].length, “y” : theArray.length }; length.max = Math.max(length.x, length.y); var temp, k, x, y; for (k = 0; k <= 2 * (length.max – 1); ++k) { for (temp = [], y = length.y – 1; (x = k … Read more
If you notice, there is a distinct pattern to the adjacency matrices you are creating. Specifically, they are symmetric and banded. You can take advantage of this fact to easily create your matrices using the diag function (or the spdiags function if you want to make a sparse matrix). Here is how you can create … Read more
I think you are mixing up an implementation detail with usage, if you will. Let’s start with a two-dimensional array, or matrix: | 1 2 3 | | 4 5 6 | | 7 8 9 | The problem is that computer memory is a one-dimensional array of bytes. To make our discussion easier, lets … Read more
The best way to solve a system of linear equations of the form Ax = b is to do the following. decompose A into the format A = M1 * M2 (where M1 and M2 are triangular) Solve M1 * y = b for y using back substitution Solve M2 * x = y for … Read more