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获取每个超级像素的RGB像素值列表

superpixels scikit-image opencv numpy python-3.x

Joseph · 技术社区 · 6 年前

l具有尺寸为(224224,3)的RGB图像。l使用SLIC算法对其进行超像素分割。

具体如下:

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)

返回的段数为:

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

从0到595。因此,我们有596个超级像素(区域)。

让我们看看 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])

我想要什么?

对于每个超级像素区域,按如下所示制作两个阵列:

1) 数组:包含属于同一超级像素的像素的索引。

例如

superpixel_list[0] 包含属于超级像素0的所有像素索引。

superpixel_list[400] 包含属于superpixel 400的像素的所有索引

2) superpixel\u pixel\u values[0]:包含像素值(在 RGB )属于超级像素0的像素数。

例如,假设像素0、24、29、53属于超级像素0。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

高效/优化的方法是什么?(因为我有l个要循环的图像数据集)

编辑-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)

在里面 superpixel_list 我们应该得到一个包含属于同一超像素的像素索引的列表。例如 superpixel\u列表[0] 应获取受影响像素的所有像素索引,以达到超级像素0

然而,我得到以下信息:

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]))

例如,在superpixel[0]中,我们应该得到受supepixel 0影响的每个像素的RGB像素值,如下所示: 例如,像素0、24、29、53受超级像素0的影响,然后:

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

但是,当我使用你的函数时,我会得到以下结果:

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])

谢谢你的帮助

1 回复 | 直到 6 年前

Daniel F 6 年前

可以使用 np.where 以及由此产生的指数。

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]