如何將圖像的所有像素值轉換為特定范圍-python

Question

我有一個12種不同顏色的rgb圖像，但我事先並不知道顏色（像素值）。 我想轉換0到11之間的所有像素值，每個像素值表示原始rgb圖像的唯一顏色。

例如，所有[230,100,140]轉換為[0,0,0]，所有[130,90,100]轉換為[0,0,1]，依此類推......所有[210,80,50]轉換為[0,0] ，11]。

Answer 1

快速而骯臟的應用。 很多都可以改進，尤其是逐像素地遍歷整個圖像並不是非常numpy也不是非常opencv，但我太懶了，不記得究竟如何閾值和替換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

你可以使用np.unique和一些np.unique ：

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_

然而，如果像素的數量遠大於顏色的數量，則以下啟發式算法可以提供巨大的加速。 它試圖找到一個廉價的哈希函數（例如只查看紅色通道），如果它成功，它使用它來創建一個查找表。 如果沒有，它會回到上述安全方法。

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)]

測試和時間：

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))

樣本輸出（12種顏色，1000x1000像素）：

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