diagonal

问题 In Matlab, I have to study the eventual existence of common eigenvectors basis between 2 Fisher matrices FISH_sp and FISH_xc of size 7x7 and diagonalisable. I get from my computation the following result: >> x=null(FISH_sp*FISH_xc-FISH_xc*FISH_sp) x = -0.0085 -0.0048 -0.2098 0.9776 -0.0089 -0.0026 0.0109 In this result, It appears that condition to get a common eigenvectors basis on commutator is true. But I need to further examine the mathematics. If one gets a single column vector, then

问题 In Matlab, I have to study the eventual existence of common eigenvectors basis between 2 Fisher matrices FISH_sp and FISH_xc of size 7x7 and diagonalisable. I get from my computation the following result: >> x=null(FISH_sp*FISH_xc-FISH_xc*FISH_sp) x = -0.0085 -0.0048 -0.2098 0.9776 -0.0089 -0.0026 0.0109 In this result, It appears that condition to get a common eigenvectors basis on commutator is true. But I need to further examine the mathematics. If one gets a single column vector, then

问题 Suppose A = np.array([[1,2,0,3,5,0,0],[5,6,7,0,9,5,10]]) (The actual data set in this case is a square matrix with a shifted diagonal of zeros) I want to create an np array V that sums the pairs reflected across the first zero in each row, in the order of their distance from the zero. i.e., V[0]=[(2+3),(1+5),0] V[1] =[(7+9),(6+5),(10+5)]... So V = ([5,6,0],[16,11,15]) I can accomplish this in a very rudimentary way by looping through each row and then adding each row to a shifted version of

问题 Say I would like to remove the diagonal from a scipy.sparse.csr_matrix . Is there an efficient way of doing so? I saw that in the sparsetools module there are C functions to return the diagonal. Based on other SO answers here and here my current approach is the following: def csr_setdiag_val(csr, value=0): """Set all diagonal nonzero elements (elements currently in the sparsity pattern) to the given value. Useful to set to 0 mostly. """ if csr.format != "csr": raise ValueError('Matrix given

问题 Say I would like to remove the diagonal from a scipy.sparse.csr_matrix . Is there an efficient way of doing so? I saw that in the sparsetools module there are C functions to return the diagonal. Based on other SO answers here and here my current approach is the following: def csr_setdiag_val(csr, value=0): """Set all diagonal nonzero elements (elements currently in the sparsity pattern) to the given value. Useful to set to 0 mostly. """ if csr.format != "csr": raise ValueError('Matrix given

问题 Say I would like to remove the diagonal from a scipy.sparse.csr_matrix . Is there an efficient way of doing so? I saw that in the sparsetools module there are C functions to return the diagonal. Based on other SO answers here and here my current approach is the following: def csr_setdiag_val(csr, value=0): """Set all diagonal nonzero elements (elements currently in the sparsity pattern) to the given value. Useful to set to 0 mostly. """ if csr.format != "csr": raise ValueError('Matrix given

来源： https://stackoverflow.com/questions/44935707/diagonal-background-structure-css

问题 I have a matrix like the following: ` matrix = [ ['P', 'o', 'P', 'o', 'P'], ['m', 'i', 'c', 's', 'r'], ['g', 'a', 'T', 'A', 'C'], ['n', 'n', 'e', 'r', 't'], ['a', 'g', 'o', 'd', 'o'], ['a', 'p', 'p', 'l', 'e'] ]` and this code which prints every letter in a 'diagonal rising to the right' way with repetitions: `test_word = '' for upper in range(len(matrix)): for rep1 in range(min(upper + 1, len(matrix[0]))): for rep2 in range(rep1, min(upper + 1, len(matrix[0]))): for j in range(rep1, rep2 + 1

问题 I have a matrix like the following: ` matrix = [ ['P', 'o', 'P', 'o', 'P'], ['m', 'i', 'c', 's', 'r'], ['g', 'a', 'T', 'A', 'C'], ['n', 'n', 'e', 'r', 't'], ['a', 'g', 'o', 'd', 'o'], ['a', 'p', 'p', 'l', 'e'] ]` and this code which prints every letter in a 'diagonal rising to the right' way with repetitions: `test_word = '' for upper in range(len(matrix)): for rep1 in range(min(upper + 1, len(matrix[0]))): for rep2 in range(rep1, min(upper + 1, len(matrix[0]))): for j in range(rep1, rep2 + 1

问题 Is there a way in Octave to compute and store only the diagonal of a matrix product? Basically like doing: vector = diag(A*B); I don't care about any of the values of A*B except those on the diagonal. The matrix sizes are around 80k x 12 and 12 x 80k , so even if I didn't care about the speed/extra memory it simply wont fit in RAM. Strange, since Octave is a package for huge data sets and diagonals are very important, so it should be possible. 回答1: The first element in the diagonal is the