python - How to define a two-dimensional array? - Stack Overflow Long back when I was not comfortable with Python, I saw the single line answers for writing 2D matrix and told myself I am not going to use 2-D matrix in Python again (Those single lines were pretty scary and It didn't give me any information on what Python was doing Also note that I am not aware of these shorthands )
python - defining the i,j th entry of a matrix in loop - Stack Overflow If your matrix is already created and thus an item already exists in row i and column j you can modify that item with A[i][j] = newvalue This works for the various "matrices" that are built in to Python, such as the list or array Note that this means that each row in the matrix is a item in the overall list, so the "matrix" is really a list
python - What does [:, :] mean on NumPy arrays - Stack Overflow numpy uses tuples as indexes In this case, this is a detailed slice assignment [0] #means line 0 of your matrix [(0,0)] #means cell at 0,0 of your matrix [0:1] #means lines 0 to 1 excluded of your matrix [:1] #excluding the first value means all lines until line 1 excluded [1:] #excluding the last param mean all lines starting form line 1 included [:] #excluding both means all lines [::2] #
matrix - How to represent matrices in python - Stack Overflow from numpy import matrix from numpy import linalg A = matrix( [[1,2,3],[11,12,13],[21,22,23]]) # Creates a matrix x = matrix( [[1],[2],[3]] ) # Creates a matrix (like a column vector) y = matrix( [[1,2,3]] ) # Creates a matrix (like a row vector) print A T # Transpose of A print A*x # Matrix multiplication of A and x print A I # Inverse of
python - Matrix indexing in Numpy - Stack Overflow I don't see how that behavior is a consequence of a matrix always being 2d In addition, A[0,:] is not the same as A[0] (even though they produce the same result in this case), so the fact that a matrix row is a matrix in the eyes of numpy does not explain the behavior I posted
python - numpy matrix vector multiplication - Stack Overflow When I multiply two numpy arrays of sizes (n x n)*(n x 1), I get a matrix of size (n x n) Following normal matrix multiplication rules, an (n x 1) vector is expected, but I simply cannot find any information about how this is done in Python's Numpy module The thing is that I don't want to implement it manually to preserve the speed of the
What is the @= symbol for in Python? - Stack Overflow @=and @ are new operators introduced in Python 3 5 performing matrix multiplication They are meant to clarify the confusion which existed so far with the operator * which was used either for element-wise multiplication or matrix multiplication depending on the convention employed in that particular library code
python - How to get element-wise matrix multiplication (Hadamard . . . matrix objects have all sorts of horrible incompatibilities with regular ndarrays With ndarrays, you can just use * for elementwise multiplication: a * b If you're on Python 3 5+, you don't even lose the ability to perform matrix multiplication with an operator, because @ does matrix multiplication now: a @ b # matrix multiplication
python - Find the linearly independent rows of a matrix - Stack Overflow from numpy import dot, zeros from numpy linalg import matrix_rank, norm def find_li_vectors(dim, R): r = matrix_rank(R) index = zeros( r ) #this will save the positions of the li columns in the matrix counter = 0 index[0] = 0 #without loss of generality we pick the first column as linearly independent j = 0 #therefore the second index is simply