使用np.tile在一批图像中平铺每个图像的10个图像

Question

Take the array: arr = [0, 1, 2]

np.tile(arr,[10,1])
array([[0, 1, 2],
       [0, 1, 2],
       [0, 1, 2],
       [0, 1, 2],
       [0, 1, 2],
       [0, 1, 2],
       [0, 1, 2],
       [0, 1, 2],
       [0, 1, 2],
       [0, 1, 2]])
>>> np.tile(arr,[10,2])
array([[0, 1, 2, 0, 1, 2],
       [0, 1, 2, 0, 1, 2],
       [0, 1, 2, 0, 1, 2],
       [0, 1, 2, 0, 1, 2],
       [0, 1, 2, 0, 1, 2],
       [0, 1, 2, 0, 1, 2],
       [0, 1, 2, 0, 1, 2],
       [0, 1, 2, 0, 1, 2],
       [0, 1, 2, 0, 1, 2],
       [0, 1, 2, 0, 1, 2]])

Similar to this, I want to use the tile function to create 10 copies of an image batch of size 10x227x227x3 (the batch already has 10 images)) For each image I want to create a tile. So I should get 100x227x227x3

However when I do this M=10):

    images = np.tile(img_batch, [M, 1])

I get 10x227x2270x3 instead, images = np.tile(img_batch, [M]) doesn't work either and brings a value of size 10x227x227x30

I can't get around my head on how to get the tiles I need. Any recommendations are welcome.

Answer 1

Your img_batch has 4 dimensions. Make the reps of size 4:

np.tile(img_batch, [M, 1, 1, 1])

Otherwise, it will be equivalent to np.tile(img_batch, [1, 1, M, 1] in your first case according to the docs :

If A.ndim > d, reps is promoted to A.ndim by pre-pending 1's to it. Thus for an A of shape (2, 3, 4, 5), a reps of (2, 2) is treated as (1, 1, 2, 2).