fastest way to find the rgb pixel color count of image

Question

I have a use case where i have to find the consecutive rgb pixel color count of each frame of live video after searching i found a piece of code which does the same thing bu

Answer 1

You need to use Numpy, or OpenCV, for fast image processing in Python. I made a 9-colour version of Paddington:

from PIL import Image
import numpy as np

# Open Paddington and make sure he is RGB - not palette
im = Image.open('paddington.png').convert('RGB')

# Make into Numpy array
na = np.array(im)

# Arrange all pixels into a tall column of 3 RGB values and find unique rows (colours)
colours, counts = np.unique(na.reshape(-1,3), axis=0, return_counts=1)

print(colours)
print(counts)

Results

[[ 14  48  84]
 [ 19  21  30]
 [ 33 108 163]
 [ 33 152 190]
 [ 72  58  58]
 [ 96 154 210]
 [180  89  64]
 [205 210 200]
 [208 151  99]]

[20389 40269 12820  1488 17185 25371 17050 16396  9032]

That means there are 20,389 pixels of RGB(14,48,84), and so on.

That takes 125ms on my Mac for a 400x400 image, which will give you 8 fps, so you better have at least 4 CPU cores and use all of them to get 25+ fps.

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Update

I think you can actually go significantly faster than this. If you take the dot-product of each of the pixels with [1,256,65536], you will get a single 24-bit number for each pixel, rather than 3 8-bit numbers. It is then a lot faster to find the unique values. That looks like this:

# Open Paddington and make sure he is RGB - not palette
im = Image.open('paddington.png').convert('RGB')

# Make into Numpy array
na = np.array(im)

# Make a single 24-bit number for each pixel
f = np.dot(na.astype(np.uint32),[1,256,65536]) 

nColours = len(np.unique(f))     # prints 9

That takes 4ms rather than 125ms on my Mac :-)

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Keywords: Python, Numpy, PIL/Pillow, image processing, count unique colours, count colors.