Question

我想显示一个矩形，其中心位于（x，y）位置，其长轴与x轴成角度phi。我怎么能这样做？

我可以通过

显示位置（x，y）的圆点

scatter(x,y)

但我不知道矩形。

Answer 1

使用fill从x和y坐标绘制2D多边形。

c = [0, 0];     % Position of centre
h = 5; w = 8;   % height and width of rectangle
a = deg2rad(30); % angle of rotation in radians (using deg2rad so can be set in degrees)
% Get corners of rectangle
corners = [c(1) + w/2, c(2) + h/2; c(1) + w/2, c(2) - h/2; c(1) - w/2, c(2) - h/2; c(1) - w/2, c(2) + h/2];
% rotate corner points
corners = [(corners(:,1)-c(1))*cos(a) - (corners(:,2)-c(2))*sin(a) + c(1), (corners(:,1)-c(1))*sin(a) + (corners(:,2)-c(2))*cos(a) + c(2)];
% Use 'fill' or 'patch' to plot
fill(corners(:,1), corners(:,2), 'r')

如果你想重用它，可以打包成一个函数

function [corners(:,1), corners(:,2)] = getRectangle(c, h, w, a)
% This function returns the corner points for a rectangle, specified
% by its centre point c = [x,y], height h, width w and angle in degrees from horizontal a
    a = deg2rad(a);
    corners = [c(1) + w/2, c(2) + h/2; c(1) + w/2, c(2) - h/2; c(1) - w/2, c(2) - h/2; c(1) - w/2, c(2) + h/2];
    corners = [(corners(:,1)-c(1))*cos(a) - (corners(:,2)-c(2))*sin(a) + c(1), (corners(:,1)-c(1))*sin(a) + (corners(:,2)-c(2))*cos(a) + c(2)];
end

使用：

[rectX, rectY] = getRectangle([0,0], 5, 8, 30);
fill(rectX, rectY, 'r');

使用散点图显示具有指定方向的矩形

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