Question

需要一个如何做到这一点的工作示例，我有一个bmp文件显示湖的形状，bmp的大小是矩形区域并且已知。我需要拍这张照片并估算湖泊的大小。到目前为止，我有一个脚本生成每个像素的巨大矩阵，告诉我它是否在湖中 - 但这不是蒙特卡洛！我需要生成随机点并以某种方式将它们与形状进行比较，这就是我遇到的问题。我不明白如何比较这里，我没有形状或线的方程，我只有确切的点信息 - 它是否在湖中。所以我想我已经有了确切的区域，但我需要找到一种比较随机点的方法。

function Yes = Point_In_Lake(x,y,image_pixel)

[pHeight,pWidth]=size(image_pixel);
    %pHeight = Height in pixel
    %pWidth = Width in pixel

width = 1000; %width is the actual width of the lake
height = 500; %height is the actual height of the lake

%converting x_value to pixel_value in image_pixel
point_x_pixel = x*pWidth/width;  
xl = floor(point_x_pixel)+1;
xu = min(ceil(point_x_pixel)+1,pWidth);

%converting y_value to pixel_value in image_pixel
point_y_pixel = y*pHeight/height;
yl = floor(point_y_pixel)+1;
yu = min(ceil(point_y_pixel)+1,pHeight);

%Finally, perform the check whether the point is in the lake 
if (image_pixel(yl,xl)~=0)&&(image_pixel(yl,xu)~=0)&&(image_pixel(yu,xl)~=0)&&(image_pixel(yu,xu)~=0)
    Yes=0;
else
    Yes=1;
end

Answer 1

以下是解决方案：

binaryMap = image_pixel

for i = 1:numel(image_pixel)

 xrand = randperm(size(image_pixel,1),1); 

 yrand = randperm(size(image_pixel,2),1); 

 Yes(i) = Point_In_Lake(xrand,yrand,binaryMap);

end

PercentLake = length(find(Yes==1))/length(Yes);

LakeArea = (PercentLake * 500000)/43560;

不规则形状的Monte Carlo模拟

1 个答案: