matrix

问题 I need to find a way to mesh triangles then to refine using refine. My vertices of my original triangles are stored inside a matrix of size nb_points * 2. My faces are stored in a nb_faces * 3 matrix. The data for each face is stored in a nb_face * 1 matrix. The meshing is done by diving the area using the middles of the segments of the triangles. Example : Origin : vertices = [0 1 ; 2 3 ; 4 1 ; 4 5]; faces = [1 2 3; 2 3 4]; data = [1 2]; Result expected after meshing : vertices = [0 1; 2 3;

问题 I want to invert matrix a but its not working. I need help. var a = math.inv([[1,2],[3,4]]) console.log(a) 回答1: Math does not contain a inv method . Therefore you cannot call it. You can however define your own matrix inversion method as is done here, or you could use the numeric JavaScript library and call it with: A = [ [1,2,3], [4,5,6], [7,3,9] ] ; Ainv = numeric.inv(A); or use the math.js library and call it with: math.inv( [ [1,2],[3,4] ] ) Notice that you need a surrounding the list of

问题 Basically, I want to reimplement this video. Given a corpus of documents, I want to find the terms that are most similar to each other. I was able to generate a cooccurrence matrix using this SO thread and use the video to generate an association matrix. Next I, would like to generate a second order cooccurrence matrix. Problem statement: Consider a matrix where the rows of the matrix correspond to a term and the entries in the rows correspond to the top k terms similar to that term. Say, k =

问题 I have a matrix of 0-1. What I would like to do is to loop into this matrix and search for the 1. Each time a 1 has been found, to simply jump or pass that row, in order to only record 1 value per row. What I am trying to know if the first episode of the sequence is a 1 . I was thinking there might be a solution with break But I am unsure how to use it properly. So this is my first matrix SoloNight = structure(c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,

问题 I have a matrix of 0-1. What I would like to do is to loop into this matrix and search for the 1. Each time a 1 has been found, to simply jump or pass that row, in order to only record 1 value per row. What I am trying to know if the first episode of the sequence is a 1 . I was thinking there might be a solution with break But I am unsure how to use it properly. So this is my first matrix SoloNight = structure(c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,

问题 I have a matrix of 0-1. What I would like to do is to loop into this matrix and search for the 1. Each time a 1 has been found, to simply jump or pass that row, in order to only record 1 value per row. What I am trying to know if the first episode of the sequence is a 1 . I was thinking there might be a solution with break But I am unsure how to use it properly. So this is my first matrix SoloNight = structure(c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0,

问题 I have a matrix (for example 100x100 dimantion): I need to do calculation with each element ( matrix[i,j]*tt/8+5 for example) I have huge matrix and I want to implement the algorithm using OpenGL shaders. I want to use shader like: uniform float val; uniform float tt; void main() { gl_Position.x = val*tt/8+5 } How I can implement the program? How I can get matrix after calculation(I do not want to show any windows\pictures? 回答1: It is possible if you create a fake window frame buffer. See my

问题 I am trying to parallelize the multiplication of two matrix A , B . Unfortunately the serial implementation is still faster than the parallel one or the speedup is too low. (with matrix dimension = 512 the speedup is like 1.3 ). Probably something is fundamentally wrong. Can someone out there give me a tip? double[][] matParallel2(final double[][] matrixA, final double[][] matrixB, final boolean parallel) { int rows = matrixA.length; int columnsA = matrixA[0].length; int columnsB = matrixB[0]

问题 I have a matrix of 1's and 0's. What I would like to do is group the cells that have 1's into clusters and count the number of clusters that exist in the matrix as well as the size of these clusters. If n number (in this case at least 4 cells with the value 1 near each other) of 1's are near each other (either immediately up, down, left or right from each other then consider them a single cluster and output the number of clusters and their size. For example the matrix looks like this: > m [,1

问题 I use Eigen library to covert matrix to quaternion, but when I turn one of the matrix to quaternion and burn it back, it turn out to be another matrix which is identity matrix. The rotation matrix I use was decomposed from a transform matrix. Eigen::Matrix3f R3d = R.topLeftCorner<3,3>(); *Rquat = R3d; R3d = (*Rquat).normalized().toRotationMatrix(); What may cause this problem? This is the matrix before change to quaternion and This is the matrix when I turn it back form the quaternion 回答1: