3

I have small image. enter image description here b g r, not gray. 

original = cv2.imread('im/auto5.png')
print(original.shape)  # 27,30,3 
print(original[13,29]) # [254 254 254]

As you can see, there is white pic (digit 14) in my image, mostly black. On the right corner (coordinates [13,29]) I get [254 254 254] - white color.

I want to calculate number of pixels with that specific color. I need it to further comparing such images with different numbers inside. There are different backgrounds on these squares, and I consider exactly white color.

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  • 1
    Welcome to Stack Overflow! What have you tried? Can you show us some code? Commented Aug 29, 2018 at 13:10
  • Do you mean you want to find all pixels matching the one at coordinates [13,29] or you want to find all pixels that are (254,254,254). What I mean is, what do you want to look for if the pixel at [13,29] is (8,7,6)? Commented Aug 29, 2018 at 13:29
  • I want to find all pixels that are (254,254,254). Not coordinates, only quanity, number of such pixels. Commented Aug 29, 2018 at 13:32

3 Answers 3

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11

I would do that with numpy which is vectorised and much faster than using for loops:

#!/usr/local/bin/python3
import numpy as np
from PIL import Image

# Open image and make into numpy array
im=np.array(Image.open("p.png").convert('RGB'))

# Work out what we are looking for
sought = [254,254,254]

# Find all pixels where the 3 RGB values match "sought", and count them
result = np.count_nonzero(np.all(im==sought,axis=2))
print(result)

Sample Output

35

It will work just the same with OpenCV's imread():

#!/usr/local/bin/python3
import numpy as np
import cv2

# Open image and make into numpy array
im=cv2.imread('p.png')

# Work out what we are looking for
sought = [254,254,254]

# Find all pixels where the 3 NoTE ** BGR not RGB  values match "sought", and count
result = np.count_nonzero(np.all(im==sought,axis=2))
print(result)
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4 Comments

Yep, was about to suggest something along those lines :)
Thanks! The way with np.count_nonzero is faster that with a cycle. One question? What is axis=2?
I timed the numpy way at 22 microseconds and the equivalent for loop at 1.5 milliseconds, so 68x faster. axis=1 runs up and down the height of the image, axis=2 runs across the image left-to-right, and axis=3 runs into the image along the R, G and B. so, by saying np.all(im==sought,axis=2) I mean that R, G and B are all equal to sought.
I've made an edit but not sure where its gone. I thought this was a great answer but wanted to point out that the OpenCV code uses BGR not RGB. Thanks
1

Image in cv2 is an iterable object. So you can just iterate through all pixels to count the pixels you are looking for.

import os
import cv2
main_dir = os.path.split(os.path.abspath(__file__))[0]
file_name = 'im/auto5.png'
color_to_seek = (254, 254, 254)

original = cv2.imread(os.path.join(main_dir, file_name))

amount = 0
for x in range(original.shape[0]):
    for y in range(original.shape[1]):
        b, g, r = original[x, y]
        if (b, g, r) == color_to_seek:
            amount += 1

print(amount)
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Comments

0

Try this :

import cv2
import numpy as np
img = cv2.imread("im/auto5.png")
color = [48,75,105] # set color to looking for
im = cv2.cvtColor(img, cv2.COLOR_BGR2RGB)
result = np.count_nonzero(np.all(im==color,axis=2))
print("Total pixels " + str(result))
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