使用散點圖顯示指定方向的矩形

Question

我想顯示一個矩形，它的中心位於（x，y）位置，其長軸與x軸成角度φ。我怎樣才能做到這一點？

我可以

scatter(x,y) 

顯示位置（X，Y）的圓形點，但我不知道該矩形的想法。

Answer 1

使用fill從其x和y座標繪製2D多邊形。

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') 

---

這可以被包裝成一個功能，如果你想重新使用它

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 

用途：

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