Question

跟进这个问题（和jorgeca的回答）： Fast Way to slice image into overlapping patches and merge patches to image我想在补丁数组的索引处添加一个偏移量，即：

A = np.arange(W*H).reshape(H,W)
P = patchify(A,(X,Y))

假设X，Y是奇数，P的大小将等于W-X + 1，H-Y + 1，因此以P [0,0]为中心的像素实际上对应于A [（Y -1）/ 2，（X-1）/ 2]。

有什么方法可以抵消（没有复制任何数据）P的指数以完美对应？

供参考，这是现有的补丁功能：

def patchify(img, patch_shape):
    img = np.ascontiguousarray(img)  # won't make a copy if not needed
    X, Y = img.shape
    x, y = patch_shape
    shape = ((X-x+1), (Y-y+1), x, y) # number of patches, patch_shape
    # The right strides can be thought by:
    # 1) Thinking of `img` as a chunk of memory in C order
    # 2) Asking how many items through that chunk of memory are needed when indices
    #    i,j,k,l are incremented by one
    strides = img.itemsize*np.array([Y, 1, Y, 1])
    return np.lib.stride_tricks.as_strided(img, shape=shape, strides=strides)

Answer 1

是以下表达式

P = np.roll(np.roll(P, X/2, 0), Y/2, 1)

你需要什么？

演示：

>>> W, H, X, Y =  10, 14, 5, 7
>>> A = np.arange(W*H).reshape(W,H)
>>> P = patchify(A,(X,Y))
>>> P = np.roll(np.roll(P, X/2, 0), Y/2, 1)
>>> all(map(all, P[X/2, Y/2] == A[:X, :Y]))
True

为Numpy数组的索引添加偏移量

1 个答案: