You have a truncated array representation. Let's look at a full example:

>>> a = np.zeros((2, 3, 4))
>>> a
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.]]])

Arrays in NumPy are printed as the word array followed by structure, similar to embedded Python lists. Let's create a similar list:

>>> l = [[[ 0.,  0.,  0.,  0.],
          [ 0.,  0.,  0.,  0.],
          [ 0.,  0.,  0.,  0.]],

          [[ 0.,  0.,  0.,  0.],
          [ 0.,  0.,  0.,  0.],
          [ 0.,  0.,  0.,  0.]]]

>>> l
[[[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]]]

The first level of this compound list l has exactly 2 elements, just as the first dimension of the array a (# of rows). Each of these elements is itself a list with 3 elements, which is equal to the second dimension of a (# of columns). Finally, the most nested lists have 4 elements each, same as the third dimension of a (depth/# of colors).

So you've got exactly the same structure (in terms of dimensions) as in Matlab, just printed in another way.

Some caveats:

1. Matlab stores data column by column ("Fortran order"), while NumPy by default stores them row by row ("C order"). This doesn't affect indexing, but may affect performance. For example, in Matlab efficient loop will be over columns (e.g. for n = 1:10 a(:, n) end), while in NumPy it's preferable to iterate over rows (e.g. for n in range(10): a[n, :] -- note n in the first position, not the last).

2. If you work with colored images in OpenCV, remember that:

   2.1. It stores images in BGR format and not RGB, like most Python libraries do.

   2.2. Most functions work on image coordinates (x, y), which are opposite to matrix coordinates (i, j).

You should use a list comprehension:

>>> import pprint
>>> n = 3
>>> distance = [[[0 for k in xrange(n)] for j in xrange(n)] for i in xrange(n)]
>>> pprint.pprint(distance)
[[[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]]]
>>> distance[0][1]
[0, 0, 0]
>>> distance[0][1][2]
0

You could have produced a data structure with a statement that looked like the one you tried, but it would have had side effects since the inner lists are copy-by-reference:

>>> distance=[[[0]*n]*n]*n
>>> pprint.pprint(distance)
[[[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]]]
>>> distance[0][0][0] = 1
>>> pprint.pprint(distance)
[[[1, 0, 0], [1, 0, 0], [1, 0, 0]],
 [[1, 0, 0], [1, 0, 0], [1, 0, 0]],
 [[1, 0, 0], [1, 0, 0], [1, 0, 0]]]

It's better performance-wise (and easier) to create a list of your 2d arrays, then to call numpy.stack():

a = []
for n in range(1, 3):
    data = np.full((2, 3), n)
    a.append(data)
a = np.stack(a)
print(a)
print(a.shape) # <- (2, 2, 3)

You can also append to the list (a) 2d arrays in a list-of-lists structure (meaning, they don't have to be numpy arrays) and call np.stack() on that "list of lists of lists", would work too.

Arrays have a fixed size. You can't append to them. If you know the upper bounds of the array, you can preallocate the array with

np.empty((10, 10, 10))

For a 10x10x10 matrix. You can then keep 3 indices x,y,z to track the actual size you have. Eg, add a new element with:

matrix[x,y,z] = newElement
x += 1

Then when you're done, you can extract the submatrix with

finalMatrix = matrix[:x,:y,:z]

a=np.empty((2,3,5))

creates a 2x3x5 array. (There is also np.zeros if you want the values initialized.)

You can also reshape existing arrays:

a=np.arange(30).reshape(2,3,5)

np.arange(30) creates a 1-d array with values from 0..29. The reshape() method returns an array containing the same data with a new shape.