查找图像的rgb像素颜色计数的最快方法 [英] fastest way to find the rgb pixel color count of image

问题描述

我有一个用例，在搜索后，我必须找到实时视频每一帧的连续rgb像素颜色计数，我发现了一段代码可以执行相同的操作，但从性能角度来看，大约需要3秒钟才能给我输出，但就我而言，我必须尽可能快地执行此计算，可能是在1秒内达到25帧.有人可以通过重构以下代码来帮助我弄清楚该怎么做

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 but performance wise it take around ~ 3 sec to give me output but in my case i have to do this calculation as fast as possible may be 25 frames in 1 seconds. Can someone help me to figure out how to do this by refactoring the below code

from PIL import Image
import timeit

starttime = timeit.default_timer()
with Image.open("netflix.png") as image:
    color_count = {}
    width, height = image.size
    print(width,height)
    rgb_image = image.convert('RGB')
    for x in range(width):
        for y in range(height):
            rgb = rgb_image.getpixel((x, y))
            if rgb in color_count:
                color_count[rgb] += 1
            else:
                color_count[rgb] = 1

    print('Pixel Count per Unique Color:')
    print('-' * 30)
    print(len(color_count.items()))
print("The time difference is :", timeit.default_timer() - starttime)

输出:

每种唯一颜色的像素数: 130869

Pixel Count per Unique Color: 130869

时差是3.9660612

The time difference is : 3.9660612

推荐答案

您需要使用Numpy或OpenCV在Python中进行快速图像处理.我制作了9色版本的Paddington:

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)

结果

[[ 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]

这意味着有20,389像素的RGB(14,48,84)，依此类推.

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

在我的Mac上，一张400x400的图片需要125毫秒的时间，这将为您提供8 fps，因此，您最好至少拥有4个CPU核心，并使用所有这些核心获得25+ fps.

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.

更新

我认为您实际上可以比这快得多.如果将每个像素的点积与[1,256,65536]相乘，则每个像素将获得一个24位数字，而不是3个8位数字.这样一来，找到唯一值的速度要快得多.看起来像这样:

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

在Mac上，这需要4毫秒而不是125毫秒:-)

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

关键字:Python，Numpy，PIL/枕头，图像处理，计算独特的颜色，计算颜色.

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

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