获取每个超像素的RGB像素值列表
Get the list of RGB pixel values of each superpixel
l 有一个尺寸为 (224,224,3) 的 RGB 图像。 l 使用SLIC算法对其进行超像素分割
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)
返回的段数是:
np.max(segments_slic)
Out[49]: 595
从 0 到 595。因此,我们有 596 个超像素(区域)。
我们来看看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?
为每个超像素区域制作两个数组如下:
1) 数组:包含属于同一超像素的像素索引。
例如
superpixel_list[0]
包含属于超像素 0 的所有像素索引。
superpixel_list[400]
包含属于超像素400
的所有像素的索引
2)superpixel_pixel_values[0] :包含属于超像素 0.
的像素的像素值(在 RGB 中)
例如,假设像素 0, 24, 29, 53 属于超像素 0。那么我们得到
superpixel[0]= [[223,118,33],[245,222,198],[98,17,255],[255,255,0]]# RGB values of pixels belonging to superpixel 0
efficient/optimized 的方法是什么? (因为我有 l 个图像数据集要循环)
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,我们应该得到一个列表,其中包含属于同一超像素的像素索引。
例如
superpixel_list[0]
应该得到受超像素 0
影响的像素的所有像素索引
但是我得到以下结果:
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]))
为什么是两个数组?
例如,在超像素[0]中,我们应该得到影响到超像素 0 的每个像素的 RGB 像素值,如下所示:
例如像素 0、24、29、53 受到超像素 0 的影响,然后:
superpixel[0]= [[223,118,33],[245,222,198],[98,17,255],[255,255,0]]
然而,当我使用你的函数时,我得到以下信息:
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])
感谢您的帮助
可以使用 np.where
和生成的索引来完成。
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]
l 有一个尺寸为 (224,224,3) 的 RGB 图像。 l 使用SLIC算法对其进行超像素分割
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)
返回的段数是:
np.max(segments_slic)
Out[49]: 595
从 0 到 595。因此,我们有 596 个超像素(区域)。
我们来看看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?
为每个超像素区域制作两个数组如下:
1) 数组:包含属于同一超像素的像素索引。
例如
superpixel_list[0]
包含属于超像素 0 的所有像素索引。
superpixel_list[400]
包含属于超像素400
2)superpixel_pixel_values[0] :包含属于超像素 0.
的像素的像素值(在 RGB 中)例如,假设像素 0, 24, 29, 53 属于超像素 0。那么我们得到
superpixel[0]= [[223,118,33],[245,222,198],[98,17,255],[255,255,0]]# RGB values of pixels belonging to superpixel 0
efficient/optimized 的方法是什么? (因为我有 l 个图像数据集要循环)
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,我们应该得到一个列表,其中包含属于同一超像素的像素索引。
例如
superpixel_list[0]
应该得到受超像素 0
但是我得到以下结果:
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]))
为什么是两个数组?
例如,在超像素[0]中,我们应该得到影响到超像素 0 的每个像素的 RGB 像素值,如下所示: 例如像素 0、24、29、53 受到超像素 0 的影响,然后:
superpixel[0]= [[223,118,33],[245,222,198],[98,17,255],[255,255,0]]
然而,当我使用你的函数时,我得到以下信息:
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])
感谢您的帮助
可以使用 np.where
和生成的索引来完成。
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]