how to make a matrix in python?

Question

Hello everyone I want to make a matrix that looks like the image, what I did first was to create a matrix of zeros and then with a for I made the diagonal of the matrix but now I need to make the diagonals that are above and below the -2 but in those there is not a single value, those have zeros and ones so I am not very clear how to make them.

try to make them with this code

N=3

D2=np.zeros((N**2,N**2))

for i in range(N**2):
  for j in range(N**2):
    if i==j:
      D2[i,j]=-2 
    elif np.abs(i-j)==1:
      D2[i,j]=1

Answer 1

You've got several choices....

1. You could "hand jam" the whole thing in by just typing the rows. (You probably already know this.)

   D = [[-2, 1, 0, ...], [...]]

2. You can make a smaller matrix that represents what you have in the red-dashed box (the sub-matrix) and just replace values in D by over-writing them with the appropriate ranges in D as such ( you could make a little loop to hit the right indices for replacement by stepping by 3...) (not shown):

---

In [30]: D
Out[30]: 
array([[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., 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.]])

In [31]: temp = [[-2, 1, 0],
    ...:         [1, -2, 1],
    ...:         [0, 1, -2]]

In [32]: D[0:3,0:3] = temp

In [33]: D
Out[33]: 
array([[-2.,  1.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 1., -2.,  1.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  1., -2.,  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.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.,  0.]])

3. You can look for a pattern in where the zeros land and augment your loop. In this case, every zero location is a total of 6 moves away from a previous zero (3 over, 3 down), so if we can anchor the first zero, we can use a modulo operator to zero out every other occurrence that is 6 moves away. The first zero is 5 moves away from the origin at (2,3) (or equivalently (3, 2) so:

---

In [34]: D = np.zeros((n,n))

In [35]: for i in range(n):
    ...:   for j in range(n):
    ...:     if i==j:
    ...:       D[i,j]=-2
    ...:     elif np.abs(i-j)==1:
    ...:       if (i+j-5) % 6 == 0:
    ...:           D[i,j]=0
    ...:       else:
    ...:           D[i,j]=1
    ...: 

In [36]: D
Out[36]: 
array([[-2.,  1.,  0.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 1., -2.,  1.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  1., -2.,  0.,  0.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0., -2.,  1.,  0.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  1., -2.,  1.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  1., -2.,  0.,  0.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0., -2.,  1.,  0.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  1., -2.,  1.],
       [ 0.,  0.,  0.,  0.,  0.,  0.,  0.,  1., -2.]])

Answer 2

One way is to build it from diagonals. Short answer

m=np.roll(np.diag(([1]*(N-1)+[0])*N), 1)-np.identity(N**2)
m+m.T

Explanation

np.diag(([1]*(N-1)+[0])*N)

is

array([[1, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 1, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 1, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 1, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0]])

If you roll it 1 (rightwise)

np.roll(np.diag(([1]*(N-1)+[0])*N), 1)

array([[0, 1, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 1, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 1, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 1, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0, 0, 1, 0],
       [0, 0, 0, 0, 0, 0, 0, 0, 1],
       [0, 0, 0, 0, 0, 0, 0, 0, 0]])

If we remove identity from that

m=np.roll(np.diag(([1]*(N-1)+[0])*N), 1)-np.identity(N**2)
m

array([[-1,  1,  0,  0,  0,  0,  0,  0,  0],
       [ 0, -1,  1,  0,  0,  0,  0,  0,  0],
       [ 0,  0, -1,  0,  0,  0,  0,  0,  0],
       [ 0,  0,  0, -1,  1,  0,  0,  0,  0],
       [ 0,  0,  0,  0, -1,  1,  0,  0,  0],
       [ 0,  0,  0,  0,  0, -1,  0,  0,  0],
       [ 0,  0,  0,  0,  0,  0, -1,  1,  0],
       [ 0,  0,  0,  0,  0,  0,  0, -1,  1],
       [ 0,  0,  0,  0,  0,  0,  0,  0, -1]])

We are halfway. We need to double diagonal, and add the symmetric one. Which we can do it easily

m+m.T

array([[-2,  1,  0,  0,  0,  0,  0,  0,  0],
       [ 1, -2,  1,  0,  0,  0,  0,  0,  0],
       [ 0,  1, -2,  0,  0,  0,  0,  0,  0],
       [ 0,  0,  0, -2,  1,  0,  0,  0,  0],
       [ 0,  0,  0,  1, -2,  1,  0,  0,  0],
       [ 0,  0,  0,  0,  1, -2,  0,  0,  0],
       [ 0,  0,  0,  0,  0,  0, -2,  1,  0],
       [ 0,  0,  0,  0,  0,  0,  1, -2,  1],
       [ 0,  0,  0,  0,  0,  0,  0,  1, -2]])