Shifting 3D numpy array

Your question is a little ill defined. First, the 'height' axis is not a very clear definition of which one you are after in a 3D array. It is much better to locate them by position in the shape tuple. I am going to take this as meaning 'along the first axis', but it should be obvious how to do it for a different one.

Second, while it is clear that you want zeros to fill in the shifted data, you don't specify what you want to do with data on the other side: should it disappear beyond the array's boundary and be lost? Or should the array be extended to keep it all?

For the former, numpy has the roll function, which does similar to what you want, but instead of filling in with zeros, it copies the data from the other side of the array. You can simply replace this with zeros afterwards:

>>> a = np.arange(60).reshape(3, 4, 5)
>>> a
array([[[ 0,  1,  2,  3,  4],
        [ 5,  6,  7,  8,  9],
        [10, 11, 12, 13, 14],
        [15, 16, 17, 18, 19]],

       [[20, 21, 22, 23, 24],
        [25, 26, 27, 28, 29],
        [30, 31, 32, 33, 34],
        [35, 36, 37, 38, 39]],

       [[40, 41, 42, 43, 44],
        [45, 46, 47, 48, 49],
        [50, 51, 52, 53, 54],
        [55, 56, 57, 58, 59]]])
>>> b = np.roll(a, 2, axis=0)
>>> b[:2,:, :] = 0
>>> b
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,  0],
        [ 0,  0,  0,  0,  0],
        [ 0,  0,  0,  0,  0],
        [ 0,  0,  0,  0,  0]],

       [[ 0,  1,  2,  3,  4],
        [ 5,  6,  7,  8,  9],
        [10, 11, 12, 13, 14],
        [15, 16, 17, 18, 19]]])

Had the shift been negative, instead of b[:shift, :, :] = 0 you would go with b[-shift:, :, :] = 0.

If you don't want to lose data, then you are basically just adding slices of zeros at the top or bottom of the array. For the first three dimensions, numpy has vstack, hstack, and dstack, and vstack should be the one you are after:

>>> a = np.arange(60).reshape(3, 4, 5)
>>> b = np.vstack((np.zeros((2,)+a.shape[1:], dtype=a.dtype), a))
>>> b
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,  0],
        [ 0,  0,  0,  0,  0],
        [ 0,  0,  0,  0,  0],
        [ 0,  0,  0,  0,  0]],

       [[ 0,  1,  2,  3,  4],
        [ 5,  6,  7,  8,  9],
        [10, 11, 12, 13, 14],
        [15, 16, 17, 18, 19]],

       [[20, 21, 22, 23, 24],
        [25, 26, 27, 28, 29],
        [30, 31, 32, 33, 34],
        [35, 36, 37, 38, 39]],

       [[40, 41, 42, 43, 44],
        [45, 46, 47, 48, 49],
        [50, 51, 52, 53, 54],
        [55, 56, 57, 58, 59]]])

If you wanted the zeros appended at the bottom, simply change the order of the parameters in the call to the stacking function.