生成高斯图像而不添加噪声

Question

我想通过Matlab生成高斯图像。它有3个圆圈（三个等级）。每个圆圈的强度之后将是高斯分布。因此，图像的直方图将是多重高斯分布为question。然而，我使用了自由噪声图像并添加了高斯噪声来制作多重高斯分布，这是非常噪声。在这个问题中，我正在寻找一种生成合成高斯图像的方法，这与我之前的方法（添加噪声）不同。感谢阅读

Answer 1

The following code creates an image following a mixture of 3 Gaussians (very easily extrapolable to more Gaussians if needed) by generating a monotonically decreasing square pattern image.

The way it work is

1. Generate random numbers following desired distributions
2. Sort said numbers
3. Spiral from center of image outwards, and insert sorted values pixel by pixel

Code:

rows=256; columns=256;
grayImage=zeros(rows,columns);

% We will work with doubles in the range of 0-255, and then we will
% discretize, for the shake of dealing properly with random numbers

%define gaussians
mean1=30;   std1=10;
mean2=100;  std2=10;
mean3=130;  std3=5;

% how many points on each of them?? 
% equal:
n=ceil(rows*columns/3);
% Random samples I tested to make it look as your image
n1=ceil(rows*columns/5);
n2=ceil(2/5*rows*columns);
n3=ceil(2/5*rows*columns);
%generate random numbers
rnd1=normrnd(mean1,std1,n1,1);
rnd2=normrnd(mean2,std2,n2,1);
rnd3=normrnd(mean3,std3,n3,1);

%now the hard part.


rnd=[rnd1;rnd2;rnd3];
% Does this looks like what you want? Tune above parameters if it doesnt.
% histogram(rnd)

% Sort the data
rnd=sort(rnd,'descend');


% Here comes the tricky part: filling the image. I chose square shaped, and
% I fill it in a spiral, starting from the center
% web('https://stackoverflow.com/questions/398299/looping-in-a-spiral')
x= 0;
y= 0;
dx = 0;
dy = -1;
next=1;
for ii= 1:rows*columns
    if (-rows/2 < x <= rows/2) && (-columns/2 < y <= columns/2)
        grayImage(x+columns/2,y+rows/2)=rnd(next);
        next=next+1;
    end
    if x == y || (x < 0 && x == -y) || (x > 0 && x == 1-y)
        auxdx=dx;
        dx=-dy; 
        dy =auxdx;
    end
    x=x+dx;
    y=y+dy;
end


% 
subplot(121);imshow(uint8(grayImage)); title('Syntetic image');
subplot(122);imhist(uint8(grayImage)); title('Histogram');

Output:

Please, do not hesitate to ask me any question about the code, but hopefully it is quite self explanatory.