获取每个超级像素的RGB像素值列表

Question

l具有尺寸的RGB图像(224,224,3(。l使用SLIC算法在其上施加了超像素分割。

如下：

img= skimageIO.imread("first_image.jpeg")
print('img shape', img.shape) # (224,224,3)
segments_slic = slic(img, n_segments=1000, compactness=0.01, sigma=1) # Up to 1000 segments
segments_slic.shape
(224,224)

返回段的数量为：

np.max(segments_slic)
Out[49]: 595

从0到595。因此，我们有596个超级像素(区域(。

让我们看一下segments_slic[0]

segments_slic[0]
Out[51]: 
array([ 0,  0,  0,  0,  0,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,  1,
        1,  2,  2,  2,  2,  2,  2,  2,  2,  2,  2,  2,  3,  3,  3,  3,  3,
        3,  3,  3,  3,  3,  4,  4,  4,  4,  4,  4,  4,  5,  5,  5,  5,  5,
        5,  5,  5,  5,  6,  6,  6,  6,  6,  6,  6,  6,  6,  7,  7,  7,  7,
        8,  8,  8,  8,  8,  8,  8,  8,  9,  9,  9,  9,  9,  9,  9,  9,  9,
       10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12,
       12, 12, 12, 12, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 14, 14, 14,
       14, 14, 14, 14, 15, 15, 15, 15, 15, 15, 15, 15, 15, 15, 16, 16, 16,
       16, 16, 16, 16, 17, 17, 17, 17, 17, 17, 17, 17, 17, 18, 18, 18, 18,
       18, 18, 18, 18, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 19, 20, 20,
       20, 20, 20, 20, 20, 20, 20, 21, 21, 21, 21, 21, 21, 21, 21, 21, 21,
       21, 21, 22, 22, 22, 22, 22, 22, 22, 22, 22, 22, 23, 23, 23, 23, 23,
       23, 23, 23, 23, 23, 23, 24, 24, 24, 24, 24, 24, 24, 25, 25, 25, 25,
       25, 25, 25])

我想得到什么？

对于每个超级像素区域，都有两个数组，如下：

1(数组：包含属于同一超像素的像素的索引。

例如

superpixel_list[0]包含属于Superpixel 0的像素的所有索引。

superpixel_list[400]包含属于Superpixel 400

的像素的所有索引

2(superpixel_pixel_values [0]：包含属于superpixel 0的像素的像素值(在 rgb 中(。

例如，假设像素0、24、29、53属于Superpixel0。然后我们得到

superpixel[0]= [[223,118,33],[245,222,198],[98,17,255],[255,255,0]]# RGB values of pixels belonging to superpixel 0

这样做的高效/优化方法是什么？(因为L具有L图像数据集可以循环(

edit-1

def sp_idx(s, index = True):
     u = np.unique(s)
         if index:
     return [np.where(s == i) for i in u]
         else:
     return [s[s == i] for i in u]
     #return [s[np.where(s == i)] for i in u] gives the same but is slower
superpixel_list = sp_idx(segments_slic)
superpixel      = sp_idx(segments_slic, index = False)

在superpixel_list中，我们应该获得一个包含属于同一超级像素的像素索引的列表。例如 superpixel_list[0]应该将所有像素的像素索引影响到Superpixel 0

但是，请列出以下内容：

superpixel_list[0]
Out[73]: 
(array([ 0,  0,  0,  0,  0,  1,  1,  1,  1,  1,  1,  2,  2,  2,  2,  2,  2,
         3,  3,  3,  3,  3,  3,  3,  4,  4,  4,  4,  4,  4,  4,  5,  5,  5,
         5,  5,  5,  5,  5,  6,  6,  6,  6,  6,  6,  6,  6,  7,  7,  7,  7,
         7,  7,  7,  8,  8,  8,  8,  8,  8,  8,  9,  9,  9,  9,  9,  9, 10,
        10, 10, 10, 10, 11, 11, 11, 11, 12, 12, 12, 12, 13, 13, 13]),
 array([0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5, 0, 1, 2, 3, 4, 5,
        6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 7, 0, 1, 2, 3, 4, 5, 6,
        7, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 6, 0, 1, 2, 3, 4, 5, 0, 1,
        2, 3, 4, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2]))

为什么要两个数组？

例如，在superpixel [0]中，我们应该将每个像素的RGB像素值对supepixel的影响到Supepixel 0如下：例如，像素0、24、29、53受到超级像素0的影响：

superpixel[0]= [[223,118,33],[245,222,198],[98,17,255],[255,255,0]]

但是，当l使用您的功能时，请列出以下内容：

superpixel[0]
Out[79]: 
array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
       0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])

谢谢您的帮助

Answer 1

可以使用np.where和结果索引完成。

def sp_idx(s, index = True):
     u = np.unique(s)
     return [np.where(s == i) for i in u]
superpixel_list = sp_idx(segments_slic)
superpixel      = [img[idx] for idx in superpixel_list]