What does .shape[] do in "for i in range(Y.shape[0])"?

The shape attribute for numpy arrays returns the dimensions of the array. If Y has n rows and m columns, then Y.shape is (n,m). So Y.shape[0] is n.

In [46]: Y = np.arange(12).reshape(3,4)

In [47]: Y
Out[47]: 
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11]])

In [48]: Y.shape
Out[48]: (3, 4)

In [49]: Y.shape[0]
Out[49]: 3

shape is a tuple that gives dimensions of the array..

>>> c = arange(20).reshape(5,4)
>>> c
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11],
       [12, 13, 14, 15],
       [16, 17, 18, 19]])

c.shape[0] 
5

Gives the number of rows

c.shape[1] 
4

Gives number of columns

shape is a tuple that gives you an indication of the number of dimensions in the array. So in your case, since the index value of Y.shape[0] is 0, your are working along the first dimension of your array.

From http://www.scipy.org/Tentative_NumPy_Tutorial#head-62ef2d3c0a5b4b7d6fdc48e4a60fe48b1ffe5006

 An array has a shape given by the number of elements along each axis:
 >>> a = floor(10*random.random((3,4)))

 >>> a
 array([[ 7.,  5.,  9.,  3.],
        [ 7.,  2.,  7.,  8.],
        [ 6.,  8.,  3.,  2.]])

 >>> a.shape
 (3, 4)

and http://www.scipy.org/Numpy_Example_List#shape has some more examples.