Another approach is through bsxfun and unique. Assuming that A is a row vector, you would do the following: un = unique(A.’, ‘stable’); out = bsxfun(@times, bsxfun(@eq, A, un), un); out contains your desired result. This code deserves some explanation. The first line of code determines all unique entries in A stored in un, but … Read more
#include <stdio.h> #include <stdlib.h> typedef int(*comparer) (int a, int b); int compareasc ( const void *pa, const void *pb ) { const int *a = *(const int **)pa; const int *b = *(const int **)pb; if(a[0] == b[0]) return a[1] – b[1]; else return a[0] – b[0]; } int comparedsc ( const void *pa, const … Read more
By the information in the question : In vector cell arrays it is usually the first vector used as a sparse matrix dimension The second element is a scalar specifying the default value of the sparse matrix The other vectors are used to specify the location and the value of the element in the sparse … Read more
Always use numpy for matrix operations. I’ll assume an m*n numpy array arr. I first did a transpose using the np.transpose function and then I flipped it using the np.fliplr function. output = np.fliplr(np.transpose(arr)) As mentioned in the comments, there is no way to do an in-place replace without a temporary variable for rectangular matrices. … Read more
Re-arranging your data into matrices with rownames makes this more convenient, as the row names are not a column, and therefore do not force a mode of character for the numeric data. Here are how the matrices would look. I created meaningless column names for this example, in order to show which columns are used … Read more
Matrix operation in R: I have a square matrix whose determinant is zero, i need to find its inverse in R programing. Is it possible ,if yes how? [closed] solved Matrix operation in R: I have a square matrix whose determinant is zero, i need to find its inverse in R programing. Is it possible … Read more
Is this what you want? mat <- as.matrix(data.frame(V1 = sample(1:2, 100, replace = TRUE), V2 = sample(1:5, 100, replace = TRUE), V3 = sample(10:50, 100, replace = TRUE), V4 = sample(1:3, 100, replace = TRUE), V5 = sample(1:5, 100, replace = TRUE))) 1 solved Creating a matrix with random generated variables in R [closed]
You were almost there. you just had to specify n and return the right object at the end. matrix_a = function(n) { A = matrix(data = 0, nrow = n, ncol = n) for (i in 1:n) { for (j in 1:n) { if (i>=j) {A[i,j] = (i^2)} if (i<j) {A[i,j] = (j^2)} } } … Read more
According to the official documentation, writelines writes a list of lines to a file. str(A) does not create a “list of lines” – obviously, when you print it to the screen – so the very first step should be to create a list-of-lines: A=[ [‘-‘, ‘A’, ‘B’, ‘C’, ‘D’, ‘E’], [‘A’, ‘0’, ‘5’, ‘6’, ‘7’, … Read more
NUMI does not have rows and columns, it is a pointer-to-pointer-to-int, and happens to be pointing at an allocated memory that has room for dim pointers-to-int, not dim * dim ints. This would be equivalent to treating as having been declared int* NUMI[dim] A call int* NUMI; NUMI= malloc( dim*dim*sizeof(int) ); will allocate a dim … Read more
int Frows = FM.length; int Fcolumns = FM[0].length; int Srows = FM[0].length; int Scolumns = FM.length; should be int Frows = FM.length; int Fcolumns = FM[0].length; int Srows = FM.length; int Scolumns = FM[0].length; and … float finAns[][] = new float[Fcolumns][Scolumns]; should be float finAns[][] = new float[Frows][Scolumns]; Start with that solved I am attempting … Read more
To do np.dot last dimension of first matrix must be the same as first dimension of second one. They are not, so you are getting ValueError, that shapes are not aligned. Everything seems to be fine as you printed, but then you forgot about lines: j_temp = np.copy(J_2_SYM) # Add row/col into J_2_SYM j_temp = … Read more
repmat and repelem are used for repeating the copies of an array. For your case, you can use either of these as: repmat(A,3,1) or repelem(A,3,1) 0 solved Matlab Repeat Value
The details will of course depend on what distribution you’ll want the matrix elements to have, but once you settle on that, you can adapt something like the following: m <- matrix(sample(1:20, 36, replace=TRUE), nrow=6) m[lower.tri(m)] <- t(m)[lower.tri(m)] m # [,1] [,2] [,3] [,4] [,5] [,6] # [1,] 19 20 15 6 5 14 # … Read more
this is to improve concept Mat A = (Mat_<float>(3, 4) << 1, 2, 3, 4, 5, 6, 7, 8, 9, 0.1, 0.1, 0.3); Mat B = (Mat_<float>(3, 4) << 1, 2, 3, 4, 5, 6, 7, 8, 9, 0.1, 0.1, 0.3); Mat AB0; multiply(A, B, AB0); cout << A << endl; cout << B << … Read more