Question

我有12种不同颜色的rgb图像，但我事先并不知道颜色（像素值）。我想转换0到11之间的所有像素值，每个像素值都表示原始rgb图像的唯一颜色。

e.g。所有[230,100,140]转换为[0,0,0]，所有[130,90,100]转换为[0,0,1]，依此类推......所有[210,80,50]转换为[0,0， 11。

Answer 1

快速而肮脏的应用程序。很多都可以改进，尤其是逐像素地遍历整个图像并不是非常numpy也不是非常开放，但我太懒了，不记得究竟如何阈值和替换RGB像素..

import cv2
import numpy as np

#finding unique rows
#comes from this answer : http://stackoverflow.com/questions/8560440/removing-duplicate-columns-and-rows-from-a-numpy-2d-array
def unique_rows(a):
    a = np.ascontiguousarray(a)
    unique_a = np.unique(a.view([('', a.dtype)]*a.shape[1]))
    return unique_a.view(a.dtype).reshape((unique_a.shape[0], a.shape[1]))

img=cv2.imread(your_image)

#listing all pixels
pixels=[]
for p in img:
    for k in p:
        pixels.append(k)

#finding all different colors
colors=unique_rows(pixels)

#comparing each color to every pixel
res=np.zeros(img.shape)
cpt=0
for color in colors:
    for i in range(img.shape[0]):
        for j in range(img.shape[1]):
            if (img[i,j,:]==color).all(): #if pixel is this color
                res[i,j,:]=[0,0,cpt] #set the pixel to [0,0,counter]
    cpt+=1

Answer 2

您可以使用np.unique进行一些诡计：

import numpy as np

def safe_method(image, k):
    # a bit of black magic to make np.unique handle triplets
    out = np.zeros(image.shape[:-1], dtype=np.int32)
    out8 = out.view(np.int8)
    # should really check endianness here
    out8.reshape(image.shape[:-1] + (4,))[..., 1:] = image
    uniq, map_ = np.unique(out, return_inverse=True)
    assert uniq.size == k
    map_.shape = image.shape[:-1]
    # map_ contains the desired result. However, order of colours is most
    # probably different from original
    colours = uniq.view(np.uint8).reshape(-1, 4)[:, 1:]
    return colours, map_

但是，如果像素数远大于颜色数，以下启发式算法可以提供巨大的加速。它试图找到一个廉价的哈希函数（例如只查看红色通道），如果它成功，它使用它来创建一个查找表。如果没有，它会回到上述安全方法。

CHEAP_HASHES = [lambda x: x[..., 0], lambda x: x[..., 1], lambda x: x[..., 2]]

def fast_method(image, k):
    # find all colours
    chunk = int(4 * k * np.log(k)) + 1
    colours = set()
    for chunk_start in range(0, image.size // 3, chunk):
        colours |= set(
            map(tuple, image.reshape(-1,3)[chunk_start:chunk_start+chunk]))
        if len(colours) == k:
            break
    colours = np.array(sorted(colours))
    # find hash method
    for method in CHEAP_HASHES:
        if len(set(method(colours))) == k:
            break
    else:
        safe_method(image, k)
    # create lookup table
    hashed = method(colours)
    # should really provide for unexpected colours here
    lookup = np.empty((hashed.max() + 1,), int)
    lookup[hashed] = np.arange(k)
    return colours, lookup[method(image)]

测试和时间安排：

from timeit import timeit

def create_image(k, M, N):
    colours = np.random.randint(0, 256, (k, 3)).astype(np.uint8)
    map_ = np.random.randint(0, k, (M, N))
    image = colours[map_, :]
    return colours, map_, image

k, M, N = 12, 1000, 1000

colours, map_, image = create_image(k, M, N)

for f in fast_method, safe_method:
    print('{:16s} {:10.6f} ms'.format(f.__name__, timeit(
        lambda: f(image, k), number=10)*100))
    rec_colours, rec_map_ = f(image, k)
    print('solution correct:', np.all(rec_colours[rec_map_, :] == image))

样本输出（12种颜色，1000x1000像素）：

fast_method        3.425885 ms
solution correct: True
safe_method       73.622813 ms
solution correct: True

如何将图像的所有像素值转换为特定范围-python

2 个答案: