Question

我正在对图像进行腐蚀。图像已相应填充。简而言之，我将一个cross元素（+）放在图像的每个像素上，并从上，下，右，左及其本身的像素中选取该像素的最小值。

效率低下，我想不出矢量化版本。由于所有计算都是彼此独立进行的，因此必须有可能。

for y in range(t,image.shape[0]-b):
    for x in range(l,image.shape[1]-r):
        a1 = numpy.copy(str_ele)
        for filter_y in range(a1.shape[0]):
            for filter_x in range(a1.shape[1]):
                if (not (numpy.isnan(a1[filter_y][filter_x]))):
                    a1[filter_y][filter_x] = str_ele[filter_y][filter_x]*image[y+(filter_y-str_ele_center_y)][x+(filter_x-str_ele_center_x)]
        eroded_image[y][x] = numpy.nanmin(a1)

基本上：

最终图像中的每个像素=距原始图像的最小值（上，下，左，右）像素

 for y in range(len(eroded_image)):
     for x in range(len(eroded_image[1])):
         eroded_image2[y][x] = numpy.nanmin(str_ele*image2[y:y+len(str_ele),x:x+(len(str_ele[1]))])

这就是我现在拥有的。还有2个循环。

Answer 1

如果image是一个填充了NaN的数组，并且您的十字形脚印正在侵蚀，您可以通过堆叠image的切片来删除for循环（以有效地向上，向左，向右和向下移动图像）然后将np.nanmin应用于切片堆栈。

import numpy as np

def orig(image):
    t, l, b, r = 1, 1, 1, 1
    str_ele = np.array([[np.nan, 1, np.nan], [1, 1, 1], [np.nan, 1, np.nan]], dtype='float')
    str_ele_center_x, str_ele_center_y = 1, 1
    eroded_image = np.full_like(image, dtype='float', fill_value=np.nan)
    for y in range(t,image.shape[0]-b):
        for x in range(l,image.shape[1]-r):
            a1 = np.copy(str_ele)
            for filter_y in range(a1.shape[0]):
                for filter_x in range(a1.shape[1]):
                    if (not (np.isnan(a1[filter_y][filter_x]))):
                        a1[filter_y][filter_x] = str_ele[filter_y][filter_x]*image[y+(filter_y-str_ele_center_y)][x+(filter_x-str_ele_center_x)]
            eroded_image[y][x] = np.nanmin(a1)   
    return eroded_image

def erode(image):
    result = np.stack([image[1:-1, 1:-1], image[2:, 1:-1], image[:-2, 1:-1], image[1:-1, 2:], image[1:-1, :-2]])
    eroded_image = np.full_like(image, dtype='float', fill_value=np.nan)
    eroded_image[1:-1, 1:-1] = np.nanmin(result, axis=0)
    return eroded_image

image = np.arange(24).reshape(4,6)
image = np.pad(image.astype(float), 1, mode='constant', constant_values=np.nan)

收益

In [228]: erode(image)
Out[228]: 
array([[nan, nan, nan, nan, nan, nan, nan, nan],
       [nan,  0.,  0.,  1.,  2.,  3.,  4., nan],
       [nan,  0.,  1.,  2.,  3.,  4.,  5., nan],
       [nan,  6.,  7.,  8.,  9., 10., 11., nan],
       [nan, 12., 13., 14., 15., 16., 17., nan],
       [nan, nan, nan, nan, nan, nan, nan, nan]])

对于上面的小示例image，erode的运行速度似乎比orig快33倍：

In [23]: %timeit erode(image)
10000 loops, best of 3: 35.6 µs per loop

In [24]: %timeit orig(image)
1000 loops, best of 3: 1.19 ms per loop

In [25]: 1190/35.6
Out[25]: 33.42696629213483

numpy向量化/更有效的循环

1 个答案: