我想计算由窗口(虚线)创建的子区域的均值和标准导数，并以识别的像素红色为中心(称为局部均值和标准导数)。这是描述它的数字
我们可以通过带掩码的卷积图像来实现。但是，这需要很长时间，因为我只关心服务器点的均值和标准推导，而卷积计算图像中的整点。你能有一种更快的方法来解决它，只计算识别像素的均值和标准导数吗？我是用matlab做的。这是我的卷积函数代码
I=[18 36 70 33 64 40 62 76 71 37 5
82 49 86 45 96 29 74 7 60 56 45
25 32 55 48 25 30 12 82 95 77 8
24 18 78 74 19 57 67 59 16 46 78
28 9 59 2 29 11 7 31 75 15 25
83 26 96 8 82 26 85 12 11 28 19
81 64 78 70 26 33 17 72 81 16 54
75 39 78 34 59 31 77 31 61 81 89
89 84 29 99 79 25 26 35 65 56 76
93 90 45 7 61 13 34 24 11 34 92
88 82 91 81 100 4 88 70 85 8 19];
identified_position=[30 36 84 90] %indices of pixel 78, 48,72 60
mask=1/9.*ones(3,3);
mean_all=imfilter(I,mask,'same');
%Mean of identified pixels
mean_all(identified_position)
% Compute the variance
std_all=stdfilt(I,ones(3));
%std of identified pixels
std_all(identified_position)
这是比较代码
function compare_mean(dimx,dimy)
I=randi(100,[dimx,dimy]);
rad=3;
identified_position=randi(max(I(:)),[1,5]);% Get 5 random position
function way1()
mask=ones(rad,rad);
mask=mask./sum(mask(:));
mean_all=conv2(I,mask,'same');
mean_out =mean_all(identified_position);
end
function way2()
box_size = rad; %// Edit your window size here (an odd number is preferred)
bxr = floor(box_size/2); %// box radius
%// Get neighboring indices and those elements for all identified positions
off1 = bsxfun(@plus,[-bxr:bxr]',[-bxr:bxr]*size(I,1)); %//'#neighborhood offsets
idx = bsxfun(@plus,off1(:),identified_position); %// all absolute offsets
I_selected_neigh = I(idx); %// all offsetted elements
mean_out = mean(I_selected_neigh,1); %// mean output
end
way2()
time_way1=@()way1();timeit(time_way1)
time_way2=@()way2();timeit(time_way2)
end
有时 way2 有错误是
Subscript indices must either be real positive integers or logicals.
Error in compare_mean/way2 (line 18)
I_selected_neigh = I(idx); %// all offsetted elements
Error in compare_mean (line 22)
way2()
最佳答案
讨论和解决方案代码
给定 I 作为输入图像，identified_position 作为所选点的线性索引，bxsz 作为窗口/框大小，接下来列出的方法必须非常有效 -
%// Get XY coordinates
[X,Y] = ind2sub(size(I),identified_position);
pts = [X(:) Y(:)];
%// Parameters
bxr = (bxsz-1)/2;
Isz = size(I);
%// XY coordinates of neighboring elements
[offx,offy] = ndgrid(-bxr:bxr,-bxr:bxr);
x_idx = bsxfun(@plus,offx(:),pts(:,1)'); %//'
y_idx = bsxfun(@plus,offy(:),pts(:,2)'); %//'
%// Outside image boundary elements
invalids = x_idx>Isz(1) | x_idx<1 | y_idx>Isz(2) | y_idx<1;
%// All neighboring indices
all_idx = (y_idx-1)*size(I,1) + x_idx;
all_idx(invalids) = 1;
%// All neighboring elements
all_vals = I(all_idx);
all_vals(invalids) = 0;
mean_out = mean(all_vals,1); %// final mean output
stdfilts = stdfilt(all_vals,ones(bxsz^2,1))
std_out = stdfilts(ceil(size(stdfilts,1)/2),:) %// final stdfilt output
bsxfun 一起获取所有已识别位置的所有相邻索引，从而获取所有这些相邻元素。然后使用这些选定的元素来获得 mean 和 stdfilt 输出。整个想法是 keep the memory requirement minimum 并同时在这些选定元素中执行 everything in a vectorized fashion 。希望这必须更快!基准测试
基准测试代码
dx = 10000; %// x-dimension of input image
dy = 10000; %// y-dimension of input image
npts = 1000; %// number of points
I=randi(100,[dx,dy]); %// create input image of random intensities
identified_position=randi(max(I(:)),[1,npts]);
rad=5; %// blocksize (rad x rad)
%// Run the approaches fed with the inputs
func1 = @() way1(I,identified_position,rad); %// original approach
time1 = timeit(func1);
clear func1
func2 = @() way2(I,identified_position,rad); %// proposed approach
time2 = timeit(func2);
clear func2
disp(['Input size: ' num2str(dx) 'x' num2str(dy) ' & Points: ' num2str(npts)])
disp(['With Original Approach: Elapsed Time = ' num2str(time1) '(s)'])
disp(['With Proposed Approach: Elapsed Time = ' num2str(time2) '(s)'])
disp(['**Speedup w/ Proposed Approach : ' num2str(time1/time2) 'x!**'])
%// OP's stated approach
function mean_out = way1(I,identified_position,rad)
mask=ones(rad,rad);
mask=mask./sum(mask(:));
mean_all=conv2(I,mask,'same');
mean_out =mean_all(identified_position);
return;
function mean_out = way2(I,identified_position,rad)
%//.... code from proposed approach stated earlier until mean_out %//
Input size: 10000x10000 & Points: 1000
With Original Approach: Elapsed Time = 0.46394(s)
With Proposed Approach: Elapsed Time = 0.00049403(s)
**Speedup w/ Proposed Approach : 939.0778x!**
关于performance - 计算图像中识别像素的局部均值和方差，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/28912312/