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]: YOut[47]: array([[ 0,  1,  2,  3],       [ 4,  5,  6,  7],       [ 8,  9, 10, 11]])In [48]: Y.shapeOut[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)>>> carray([[ 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.

Fromhttp://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 moreexamples.