如何检查 RGB colors 列表中的所有项目是否在没有循环的图像中？

Question

Say I have a list of RGB values as follows:

rgbL = [[20 45 40] [30 45 60] .... [70 50 100]]

Then, I have an image say, img = cv.imread("location")

Now, I want to change ALL RGB values of an image into (255, 0, 0) if the image RGB value is IN my list of RGB values (rgbL).

I was able to do this by this code:

for rgb in rgbL :

    k = list(filter(None, rgb[1:-1].split(" ")))
    r = int(k[0])
    g = int(k[1])
    b = int(k[2])

    img[np.all(img == (r, g, b), axis=-1)] = (255,0,0)  

But the code above is taking too long because my "rgbL" list is too long.

Is there a way that I can do this without a loop? What is the best way to implement it in numpyish way?

Answer 1

convert your rgbL and img into numpy arrays. one way of doing it without loop:

sh = img.shape
img = img.reshape(-1, 3)
img[np.where(((rgbL[:,None,:]-img)==0).all(axis=2))[1]]=np.array([255,0,0])
img = img.reshape(sh)

which takes a difference of your image with every row of rgbL and checks for all zero difference in RGBs to replace using np.where .

sample img and output:

img:
[[ 20  45  40]
 [ 30  45  60]
 [  0   1   2]
 [ 70  50 100]
 [  4   5   6]]
rgbL:
[[ 20  45  40]
 [ 30  45  60]
 [ 70  50 100]]
Output:
[[255   0   0]
 [255   0   0]
 [  0   1   2]
 [255   0   0]
 [  4   5   6]]

UPDATE : Per OP's comment on converting string dict keys to numpy arrays:

rgbL = np.array([list(map(int,[s.strip() for s in key.strip('[').strip(']').strip(' ').split(' ') if s.strip()])) for key in rgb_dict.keys()])

Answer 2

The algorithm mentioned above is O(length of rgbL * size of image)

We can bring it down to O(length of the rgbL + size of image)

set_rgbl = set(rgbL)
height = img.shape[0]
width = img.shape[1]
BLUE_COLOR = (255, 0, 0)

for h in range(0, height):
    for w in range(0, width):
        if list(img[h, w]) in set_rgbl:
            img[h, w] = BLUE_COLOR

What happens here is that we created a set of rgbL values. The insert in set is constant time operation. Same holds for lookup. Hence, when we are iterating over every pixel values in image for every pixel we spend O(1) time. This leads to improvement in our time complexity.

Hope that helps.