Python: Appending dimensions to a bidimensional array

Suppose you have an array (m, m) and want to make it (n, n). For example, transforming a 2x2 matrix to a 6x6. So:

[[ 1.  2.]
 [ 3.  4.]]

To:

[[ 1.  2.  0.  0.  0.  0.]
 [ 3.  4.  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.]]

This is what I'm doing:

def array_append(old_array, new_shape):
    old_shape = old_array.shape
    dif = np.array(new_shape) - np.array(old_array.shape)
    rows = []
    for i in xrange(dif[0]):
        rows.append(np.zeros((old_array.shape[0])).tolist())
    new_array = np.append(old_array, rows, axis=0)
    columns = []
    for i in xrange(len(new_array)):
        columns.append(np.zeros(dif[1]).tolist())
    return np.append(new_array, col开发者_运维知识库umns, axis=1)

Example use:

test1 = np.ones((2,2))
test2 = np.zeros((6,6))
print array_append(test1, test2.shape)

Output:

[[ 1.  1.  0.  0.  0.  0.]
 [ 1.  1.  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.]]

Based on this answer. But that's a lot of code for an (imho) simple operation. Is there a more concise/pythonic way to do it?

---

Why not use array = numpy.zeros((6,6)), see the numpy docs...

EDIT, woops, question has been edited... I guess you are trying to put ones in a section of an array filled with zeros? Then:

array = numpy.zeros((6,6))
array[0:2,0:2] = 1

If the small matrix does not all have the value of 1:

array[ystart:yend,xstart:xend] = smallermatrix

---

That would be then:

# test1= np.ones((2, 2))
test1= np.random.randn((2, 2))
test2= np.zeros((6, 6))
test2[0: 2, 0: 2]= test1