代码:
shape = np.array([6, 6])
grid = np.array([x.ravel() for x in np.meshgrid(*[np.arange(x) for i, x in enumerate(shape)], indexing='ij')]).T
slices = [tuple(slice(box[i], box[i] + 2) for i in range(len(box))) for box in grid]
score = np.zeros((7,7,3))
column = np.random.randn(36, 12) #just for example
column
>> array([[ 0, 1, 2, 3, ... 425, 426, 427, 428, 429, 430, 431]])
column = column.reshape((16, 3, 3, 3))
for i, window in enumerate(slices):
score[window] += column[i]
score
>> array([[[0.000e+00, 1.000e+00, 2.000e+00],
[3.000e+01, 3.200e+01, 3.400e+01],
[9.000e+01, 9.300e+01, 9.600e+01], ...
[8.280e+02, 8.300e+02, 8.320e+02],
[4.290e+02, 4.300e+02, 4.310e+02]]])
它可以工作,但是最后两行要花很多时间,因为它们会循环播放。问题是“ grid”变量包含一个窗口数组。而且我现在不知道如何加快这一过程。
答案 0 :(得分:2)
让我们从根本上简化问题-减小尺寸,并减小最终尺寸3尺寸:
In [265]: shape = np.array([4,4])
In [266]: grid = np.array([x.ravel() for x in np.meshgrid(*[np.arange(x) for i
...: , x in enumerate(shape)], indexing='ij')]).T
...: grid = [tuple(slice(box[i], box[i] + 3) for i in range(len(box))) fo
...: r box in grid]
...:
...:
In [267]: len(grid)
Out[267]: 16
In [268]: score = np.arange(36).reshape(6,6)
In [269]: X = np.array([score[x] for x in grid]).reshape(4,4,3,3)
In [270]: X
Out[270]:
array([[[[ 0, 1, 2],
[ 6, 7, 8],
[12, 13, 14]],
[[ 1, 2, 3],
[ 7, 8, 9],
[13, 14, 15]],
[[ 2, 3, 4],
[ 8, 9, 10],
[14, 15, 16]],
....
[[21, 22, 23],
[27, 28, 29],
[33, 34, 35]]]])
这是一个移动的窗口-一个(3,3)数组,移1,...,移1,等等
使用as_strided
可以构造由所有这些窗口组成的数组视图,但实际上不复制值。在与as_strided
合作之前,我就可以建立如下的等效步幅:
In [271]: score.shape
Out[271]: (6, 6)
In [272]: score.strides
Out[272]: (48, 8)
In [273]: ast = np.lib.stride_tricks.as_strided
In [274]: x=ast(score, shape=(4,4,3,3), strides=(48,8,48,8))
In [275]: np.allclose(X,x)
Out[275]: True
这可以扩展到您的(28,28,3)维度,并变成总和。
生成此类移动窗口已在先前的SO问题中进行了介绍。而且还可以在其中一个图像处理包中实现。
适应3通道图像
In [45]: arr.shape
Out[45]: (6, 6, 3)
In [46]: arr.strides
Out[46]: (144, 24, 8)
In [47]: arr[:3,:3,0]
Out[47]:
array([[ 0., 1., 2.],
[ 6., 7., 8.],
[12., 13., 14.]])
In [48]: x = ast(arr, shape=(4,4,3,3,3), strides=(144,24,144,24,8))
In [49]: x[0,0,:,:,0]
Out[49]:
array([[ 0., 1., 2.],
[ 6., 7., 8.],
[12., 13., 14.]])
由于我们一次将窗口移动一个元素,因此x
的步幅很容易从源步幅中得出。
对于4x4窗口,只需更改形状
x = ast(arr, shape=(3,3,4,4,3), strides=(144,24,144,24,8))
在Efficiently Using Multiple Numpy Slices for Random Image Cropping
@Divikar建议使用skimage
使用默认的step=1
,结果兼容:
In [55]: from skimage.util.shape import view_as_windows
In [63]: y = view_as_windows(arr,(4,4,3))
In [64]: y.shape
Out[64]: (3, 3, 1, 4, 4, 3)
In [69]: np.allclose(x,y[:,:,0])
Out[69]: True