Is there a numpy function like np.fill(), but for arrays as fill value?

问题

I'm trying to build an array of some given shape in which all elements are given by another array. Is there a function in numpy which does that efficiently, similar to np.full(), or any other elegant way, without simply employing for loops?

Example: Let's say I want an array with shape (dim1,dim2) filled with a given, constant scalar value. Numpy has np.full() for this:

my_array = np.full((dim1,dim2),value)

I'm looking for an analog way of doing this, but I want the array to be filled with another array of shape (filldim1,filldim2) A brute-force way would be this:

my_array = np.array([])
for i in range(dim1):
    for j in range(dim2):
        my_array = np.append(my_array,fill_array)
my_array = my_array.reshape((dim1,dim2,filldim1,filldim2))

EDIT

I was being stupid, np.full() does take arrays as fill value if the shape is modified accordingly:

my_array = np.full((dim1,dim2,filldim1,filldim2),fill_array)

Thanks for pointing that out, @Arne!

回答1:

You can use np.tile:

>>> shape = (2, 3)
>>> fill_shape = (4, 5)
>>> fill_arr = np.random.randn(*fill_shape)

>>> arr =  np.tile(fill_arr, [*shape, 1, 1])

>>> arr.shape
(2, 3, 4, 5)

>>> np.all(arr[0, 0] == fill_arr)
True

---

Edit: better answer, as suggested by @Arne, directly using np.full:

>>> arr = np.full([*shape, *fill_shape], fill_arr)                                               

>>> arr.shape                                                                                    
(2, 3, 4, 5)

>>> np.all(arr[0, 0] == fill_arr)                                                                
True

来源：https://stackoverflow.com/questions/62771314/is-there-a-numpy-function-like-np-fill-but-for-arrays-as-fill-value