How to extract the specific row and make that into new matrix?

Question

Here is my question.

There is a matrix call 'dt'.

What I want to do is make new matrix with same dt[:,0].

Below example will be helpful to understand what I want to do.

ex.

dt = [[0,0,0,0]
      [0,0,1,444]
      [0,0,2,80]
      [0,0,3,5]
      [1,0,0,0]
      [1,0,1,444]
      [1,0,2,80]
      [1,0,4,75]
      [2,1,2,653]
      ...
      ]]


new_matrix_0 =
     [[0,0,0,0]
      [0,0,1,444]
      [0,0,2,80]
      [0,0,3,5]]

new_matrix_1 = 
    [[1,0,0,0]
     [1,0,1,444]
     [1,0,2,80]
     [1,0,4,75]]

I need a code between 'dt' >> 'new_matrix_()'.

Thanks.

Answer 1

You can just loop through dt and append the new matrix to a list.

def get_new_matrices(dt):
    new_matrices = []
    n_rows = len(dt)
    i = 0
    while i < n_rows:
        new_matrices.append(dt[:][i:i+4])
        i = i + 4
    new_matrices.append(dt[:][i-4:n_rows])
    return new_matrices


dt = [[0, 1, 2, 3],
      [4, 5, 6, 7],
      [0, 1, 2, 3],
      [4, 5, 6, 7],
      [0, 1, 2, 3],
      [4, 5, 6, 7],
      [0, 1, 2, 3],
      [4, 5, 6, 7]]
new_mats = get_new_matrices(dt)

You can also use np.array_split

dt = [[0, 1, 2, 3],
      [4, 5, 6, 7],
      [0, 1, 2, 3],
      [4, 5, 6, 7],
      [0, 1, 2, 3],
      [4, 5, 6, 7],
      [0, 1, 2, 3],
      [4, 5, 6, 7]]
n = len(dt)
x = np.array(dt)
new_mats = np.array_split(x, int(n/4))