Construct 3 dimensions array from two dimensions array using tile

问题

I have an array arr = np.array([0.0, 0.0, 1.0])

I want to contract array of following shape promo_class.shape which is (2,3,3)

I want to create repeated array of shape (2,3,3)

array([[[0., 0., 1.],
        [0., 0., 1.],
        [0., 0., 1.]],
       [[0., 0., 1.],
        [0., 0., 1.],
        [0., 0., 1.]]])

Any idea how to do that with np.tile function ?

回答1:

Simply use np.broadcast_to for a view -

In [142]: arr = np.array([0.0, 0.0, 1.0])

In [144]: np.broadcast_to(arr,(2,3,3))
Out[144]: 
array([[[0., 0., 1.],
        [0., 0., 1.],
        [0., 0., 1.]],

       [[0., 0., 1.],
        [0., 0., 1.],
        [0., 0., 1.]]])

Why should we use view?

Because being a view, it has no extra memory overhead and hence virtually free on runtime -

In [148]: arr = np.random.rand(300)

In [149]: %timeit np.broadcast_to(arr,(200,300,300))
100000 loops, best of 3: 3.13 µs per loop

If you need an output with its own memory space, append with .copy().

---

If you are devoted to np.tile -

In [174]: np.tile(arr,(2,3,1))
Out[174]: 
array([[[0., 0., 1.],
        [0., 0., 1.],
        [0., 0., 1.]],

       [[0., 0., 1.],
        [0., 0., 1.],
        [0., 0., 1.]]])

来源：https://stackoverflow.com/questions/55607273/construct-3-dimensions-array-from-two-dimensions-array-using-tile