WebFor the specialized case of matrices, a simple slicing is WAY faster then numpy.kron() (the slowest) and mostly on par with numpy.einsum()-based approach (from @Divakar answer).Compared to scipy.linalg.block_diag(), it performs better for smaller arr, somewhat independently of number of block repetitions.. Note that the performances of … WebAug 20, 2015 · A.diagonal is a method of numpy.ndarray, just as the print out suggests. Therefore, the solution of @Saullo Castro works for numpy arrays as well, without the need to convert to np.matrix. import numpy as np A = np.arange (25).reshape ( (5,5)) diag = A.diagonal () # array ( [ 0, 6, 12, 18, 24])
numpy.chararray.diagonal — NumPy v1.4 Manual (DRAFT)
WebMay 23, 2024 · You can use numpy.diag_indices_from () to get the indices of the diagonal elements of your array. Then set the value of those indices. X [np.diag_indices_from (X)] = 0. Example: WebMay 27, 2015 · Here is a solution for a constant tri-diagonal matrix, but my case is a bit more complicated than that. I know I can do that with a loop or with list comprehension, but are there other ways? ... Make special diagonal matrix in Numpy. Related. 225. Create a list with initial capacity in Python. 762. onondaga county iqs
numpy.matrix.diagonal — NumPy v1.4 Manual (DRAFT)
WebConstruct a sparse matrix from diagonals. Parameters: diagonalssequence of array_like Sequence of arrays containing the matrix diagonals, corresponding to offsets. offsetssequence of int or an int, optional Diagonals to set: k = 0 the main diagonal (default) k > 0 the kth upper diagonal k < 0 the kth lower diagonal shapetuple of int, optional WebApr 4, 2010 · Diagonal values are left untouched. a -- square NumPy array, such that a_ij = 0 or a_ji = 0, for i != j. """ return a + a.T - numpy.diag (a.diagonal ()) This works under reasonable assumptions (such as not doing both a [0, 1] = 42 and the contradictory a [1, 0] = 123 before running symmetrize ). WebThe range # is -x+1 to y (exclusive of y), so for a matrix like the example above # (x,y) = (4,5) = -3 to 4. diags = [a[::-1,:].diagonal(i) for i in range(-a.shape[0]+1,a.shape[1])] # Now back to the original array to get the upper-left-to-lower-right diagonals, # starting from the right, so the range needed for shape (x,y) was y-1 to -x+1 ... in windows system32 spool printers