Question

我想在2D笛卡尔坐标系中绘制一个正方形，其角点位于(±1,±1)。我想进一步将其划分为400个较小且相等的正方形，每个正方形的边长为0.1。

我怎样才能在MATLAB中做到这一点？

Answer 1

您可以生成具有正确数量的垂直和水平线的网格：

%%
N = 400;
x = linspace(-1,1,sqrt(N)+1)
y = linspace(-1,1,sqrt(N)+1)

% Horizontal grid 
for k = 1:length(y)
  line([x(1) x(end)], [y(k) y(k)])
end

% Vertical grid
for k = 1:length(y)
  line([x(k) x(k)], [y(1) y(end)])
end

axis square

Answer 2

这看起来像我必须解决的问题。我在下面做的是使用meshgrid获取所有点的坐标。然后我通过pdist得到从everey点到每个其他点的距离，当距离为1时，它是我们想要绘制的连接。然后我们绘制所有这些线。

%# enter your prerequisites
I=400; R=0.1; N=sqrt(I); M=sqrt(I);

%# create vertices on square grid defined by two vectors
[X Y] = meshgrid(1:N,1:M); X = X(:); Y = Y(:); 

%# create adjacencymatrix with connections between all neighboring vertices
adjacency = squareform( pdist([X Y], 'cityblock') == 1 );

%# plot adjacenymatrix on grid with scale R and origin at the center
[xx yy] = gplot(adjacency, [X Y]); 
xx = xx-round(sqrt(I)/2); %# this centers the origin
yy = yy-round(sqrt(I)/2);
plot(xx*R, yy*R)

Answer 3

请参阅rectangle功能。例如，尝试

% Draw large bounding box:
xstart = -1;
ystart = -1;

xlen = 2;
ylen = 2;

rectangle('position', [xstart, ystart, xlen, ylen])

% Draw smaller boxes
dx = 0.1;
dy = 0.1;

nx = floor(xlen/dx);
ny = floor(ylen/dy);

for i = 1:nx
    x = xstart + (i-1)*dx;
    for j = 1:ny
        y = ystart + (j-1)*dy;
        rectangle('position', [x, y, dx, dy])
    end
end

在MATLAB中绘制正方形网格

3 个答案: