How to create a matrix?

How to create a Matrix through an iteration in Python

• I want to create a (3n*7) Matrix in python through Iteration, I did this in VB.Net and it worked but it gives me some challenge in Python as this a new language to me. n can be the number of Iterations to build up the matrix, i.e. ; if u iterated 3 times, the matrix will be a 9*7 matrix. Here is how I did it in VB.Net. : 'Assign some values to the A matrix A_mat(p, 0) = 1 : A_mat(1 + p, 1) = 1 : A_mat(2 + p, 2) = 1 'row/column 1,2 & 3 data A_mat(p, 3) = c : A_mat(p, 4) = 0 : A_mat(p, 5) = a : A_mat(p, 6) = b 'row 1 data A_mat(1 + p, 3) = g : A_mat(1 + p, 4) = d : A_mat(1 + p, 5) = t : A_mat(1 + p, 6) = f 'row 2 data A_mat(2 + p, 3) = m : A_mat(2 + p, 4) = h : A_mat(2 + p, 5) = u : A_mat(2 + p, 6) = k 'row 3 data this yielded: 1 0 0 c 0 a b 0 1 0 g d t f 0 0 1 m h u k . . . . . .

• Answer:

  One solution is to use a numpy (http://new.scipy.org/download.html) matrix: >>> from numpy import zeros, matrix >>> n = 2 >>> mat = matrix(zeros([3*n, 7])) >>> mat matrix([[ 0., 0., 0., 0., 0., 0., 0.], [ 0., 0., 0., 0., 0., 0., 0.], [ 0., 0., 0., 0., 0., 0., 0.], [ 0., 0., 0., 0., 0., 0., 0.], [ 0., 0., 0., 0., 0., 0., 0.], [ 0., 0., 0., 0., 0., 0., 0.]]) However, as noted above by @joris, with a numpy matrix all of your data must be of the same type. >>> mat[0,2] = 14 >>> mat matrix([[ 0., 0., 14., 0., 0., 0., 0.], [ 0., 0., 0., 0., 0., 0., 0.], [ 0., 0., 0., 0., 0., 0., 0.], [ 0., 0., 0., 0., 0., 0., 0.], [ 0., 0., 0., 0., 0., 0., 0.], [ 0., 0., 0., 0., 0., 0., 0.]]) >>> mat[0,1] = 's' Traceback (most recent call last): File "<pyshell#139>", line 1, in <module> mat[0,1] = 's' ValueError: could not convert string to float: s You probably want to use a nested list instead, since this is by far a more flexible and generic data type, and in addition is easier to get the hang of. For that I would refer you to http://stackoverflow.com/questions/2397141/how-to-initialize-a-two-dimensional-array-in-python, since @Mike Graham does a right proper job of explaining them. You might want to define a method like: def shape_zeros(n): # Here's where you build up your "matrix"-style nested list pass # The pass statement does nothing! Here are a couple of hints for filling up the code body of the above function. First, you can construct a whole list of duplicates with the * asterisk operator: >>> row = [0] * 7 >>> row [0, 0, 0, 0, 0, 0, 0] Next, you can nest lists within lists; but be careful when moving lists around, as the list's name is actually like a pointer: >>> mat = [] >>> n = 2 >>> for i in range(n*3): mat.append(row) >>> mat [[0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0]] >>> mat[0][0] = 1 >>> mat [[1, 0, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0, 0], [1, 0, 0, 0, 0, 0, 0]] You can avoid the above problem by creating your sub-lists (rows) separately or using a list(old_list) constructor function to make a copy. When you build it right, you can access/manipulate an element of a nested list like this: >>> mat[0][0] = 1 >>> mat[1][2] = 'Q_Q' >>> mat[2][0] = 3 >>> mat[2][2] = 5 >>> mat [[1, 0, 0, 0, 0, 0, 0], [0, 0, 'Q_Q', 0, 0, 0, 0], [3, 0, 5, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0, 0]] Good luck!