计算特定坐标下每个RGB通道均值的最快方法

Question

我为此构建了很多函数，最快的一个是:

meanRGB[:] = sum([img[i,j,:] for i,j in args])/len(args)

但是对于大图像来说仍然很慢。一张4700x3100的图片大约需要40秒。(它运行在一个旧的cpu(第四代)，但这并不重要)

参数包含特定x,y的二维坐标

[[0 0] [1 0] [2 1] ...]

感谢所有发送这些伟大方法的人。我觉得joostblack的方法是最快的。您可以在这里模拟这种情况(缩写):

clusterMap = [[1,4,2],[1,2,3],[3,1,4]]
np.random.seed(2)
img = np.random.randint(low = 0, high = 255, size = (4700, 3100, 3))
segments = np.unique(clusterMap)
meanRGB = np.zeros((len(segments),3))
for k, segment in enumerate(segments.ravel()):
args = np.argwhere(clusterMap==segment)
meanRGB[k,:] = sum([img[i,j,:] for i,j in args])/len(args)

print(f"n{meanRGB}")    

Answer 1

使用numba:

import numpy as np
from numba import njit, prange
img = np.random.randint(0, 256, (4700, 3100, 3))
args = np.random.randint(0, 1000, (100000, 2))
@njit(parallel=True, fastmath=True)
def average(array, indices):
n = len(indices)
sum_ = np.zeros(3)
for i in prange(n):
sum_ += array[indices[i][0], indices[i][1], :]
return sum_/n

验证:

op = sum(img[i,j,:] for i,j in args)/len(args)
numba_ = average(img, args) 
np.isclose(op, numba_)
# array([ True,  True,  True])

基准测试(大约快几百倍):

%%timeit
meanRGB = sum(img[i,j,:] for i,j in args)/len(args)
# 124 ms ± 704 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
%timeit average(img, args)
# 136 µs ± 12.1 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)

Answer 2

我试了一下。我认为这是更快的，但我没有你的图像或你的args数组，所以我不确定。

import numpy as np
# Trying to recreate your situation
img = np.random.rand(4700,3100,3)
args = np.array([[2,3],[0,1],[2,4],[5,5]])
# my attempt
meanRGB = np.mean(img[args[:,0],args[:,1],:],axis=0)
print(meanRGB)

(例子)输出:

[0.68078292 0.4485302  0.46863616]

Answer 3

按通道分割，然后使用内置平均值

r = np.mean(img[:,:,0])
g = np.mean(img[:,:,1])
b = np.mean(img[:,:,2])

如果你的列表不是完整的图像，你想用

遮罩它

mask = np.zeros_like(test[:,:,0])
x_coords = [c[0] for c in args]
y_coords = [c[1] for c in args]
mask[x_coords,y_coords] = 1
r = np.mean(test[mask==1][:,0])
g = np.mean(test[mask==1][:,1])
b = np.mean(test[mask==1][:,2])

Answer 4

您可以按以下方式操作:

red_channel_mean = img[:, :, 0].mean()
green_channel_mean = img[:, :, 1].mean()
blue_channel_mean = img[:, :, 2].mean()

欢呼

Answer 5

您可以将索引压缩在一起，屏蔽输入并计算平均值:

>>> img[tuple(zip(*args))].mean(0)

例如，给定args为

>>> args
[[1, 0], [2, 1], [1, 4], [0, 2]]
>>> tuple(zip(*args))
((1, 2, 1, 0), (0, 1, 4, 2))