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pythonimagecolorsaverage

Python - average color of part of an image

I need to find the average color of a specified part of the image. Specifically I need to know If the color in there is red, green, brown, black or blue.

The images are 640x640, and the region I need to analyze is the rectangle between pixel x1=20 y1=600 and pixel x2=620 y2=640. (where x0y0 is top left corner)

I found several examples, like this one How to find the average colour of an image in Python with OpenCV? but they all deal with the whole image.

How can I get the average color of only apecified area?

A must is that it has to be as quick as possible, below 5 ms.

my aproach would be to go over each pixel in the range and do the maths, but I have the feeling that open CV or similar libraries already can do this.

Any help appreciated.

Solution

  • As your region of interest (ROI) is only a simple rectangle, I think you just want to use Numpy slicing to identify it.

    So, I have made a test image that is green where you want to measure:

    enter image description here

    Then the code would go like this:

    import cv2
import numpy as np

# Load the image
im = cv2.imread('start.png')

# Calculate mean of green area
A = np.mean(im[600:640, 20:620], axis=(0,1))

    That gets green, unsurprisingly:

    array([  0., 255.,   0.])

    Now include some of the black area above the green to reduce the mean "greenness"

    B = np.mean(im[500:640, 20:620], axis=(0,1))

    That gives... "a bit less green":

    aarray([ 0.        , 72.85714286,  0.        ])

    The full sampling of every pixel in the green area takes 214 microsecs on my Mac, as follows:

    IIn [5]: %timeit A = np.mean(im[600:640, 20:620], axis=(0,1))
214 µs ± 150 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)

    Note that you could almost certainly sample every 4th pixel down and every 4th pixel across as follows in 50.6 microseconds and still get a very indicative result:

    In [11]: %timeit A = np.mean(im[500:640:4, 20:620:4], axis=(0,1))
50.6 µs ± 29.3 ns per loop (mean ± std. dev. of 7 runs, 10000 loops each)

    You can make every pixel you are sampling into a red dot like this - look carefully:

    im[600:640:4, 20:620:4] = [255,0,0]

    enter image description here

    As suggested by Fred (@fmw42), it is even faster if you replace np.mean() with cv2.mean():

    So, 11.4 microseconds with cv2.mean() versus 214 microseconds with np.mean():

    In [22]: %timeit cv2.mean(im[600:640, 20:620])
11.4 µs ± 11.8 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

    And 7.85 microseconds with cv2.mean() versus 50.6 microseconds with np.mean() if sampling every 4th pixel:

    In [23]: %timeit cv2.mean(im[600:640:4, 20:620:4])
7.85 µs ± 6.42 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)