I had a try at this...
First off, I noticed that in the following table of RGB values, the Green values are all the same so there is no point checking them.
Secondly, if you divide the values in the array by 255, you get zeroes and ones which are very close to the labelling you need. So, if you do a little maths:
t = R/255 + 2B/255 -1
then you get this for the values in the dictionary:
R G B t
==================
255 255 0 0
0 255 255 1
255 255 255 2
The code to compare with a couple of other answers looks like this:
#!/usr/bin/env python3
import numpy as np
def me(img):
"""Return R + 2B - 1 as label"""
return np.uint8((img[:,:,0]/255) + 2*(img[:,:,2]/255) - 1)
def deepak(img):
r = np.array([255, 255, 0])
g = np.array([0, 255, 255])
b = np.array([255, 255, 255])
label_seg = np.zeros((img.shape[:2]), dtype=np.uint8)
label_seg[(img==r).all(axis=2)] = 0
label_seg[(img==g).all(axis=2)] = 1
label_seg[(img==b).all(axis=2)] = 2
return label_seg
def marios(label):
mask_mapping = {
(255, 255, 0): 0,
(0, 255, 255): 1,
(255, 255, 255): 2,
}
for k in mask_mapping:
label[(label == k).all(axis=2)] = mask_mapping[k]
return label
# Generate a sample image
img = np.zeros((480,640,3), dtype=np.uint8)
img[:160,:,:] = [255,255,0]
img[160:320,:,:] = [0,255,255]
img[320:,:,:] = [255,255,255]
The timings come out like this:
In [134]: %timeit deepak(img)
15.4 ms ± 181 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [135]: %timeit marios(img)
15.4 ms ± 166 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In [172]: %timeit me(img)
869 µs ± 8.93 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)each)
Whether the 18x speedup is worth the less readable code is another argument, though comments can go a good way to helping :-)
Note, in fairness to Deepak, his time can be reduced to 0 10.3ms by removing the unnecessary line below which zeroes some elements in an array of zeroes:
label_seg[(img==r).all(axis=2)] = 0