How to create a transparent, rectangular patch with rounded corners?

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Chris L'Esperance
Chris L'Esperance 2019-1-16
评论： jon erickson 2024-5-15
It would be helpful to create a transparent, rectangular patch with rounded corners. This could be acheived with something like this:
h = rectangle('Position', [x1, y1, x2, y2], ...
'Curvature', 0.2, ...
'FaceColor', 'r', ...
'EdgeColor', 'r');
Unfortunately, rectangle properties do not include FaceAlpha and EdgeAlpha. Patch objects do, so I thought that I could get the XData and YData properties of the rectangle object from which to create a patch object. No dice; wasn't able to find these properties.
Again, the desired features are: transparency + rounded corners + face color

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采纳的回答

Afiq Azaibi
Afiq Azaibi 2024-4-26
移动：Adam Danz 2024-5-8
Ran in:
Starting in R2024a, rectangle supports controling its transparency using the FaceAlpha property:
bar(magic(4));
r = rectangle(Position=[1.5,0,1,12],FaceColor='r',FaceAlpha=.3,Curvature=[.4 .4]);

更多回答（5 个）

Jon
Jon 2019-9-18
You can use the rectangle function if you give the color as a triple instead of a string and add a fourth argument which specifices the alpha. Here, the alpha is 0.7.
h = rectangle('Position', [x1, y1, x2, y2], ...
'Curvature', 0.2, ...
'FaceColor', [1, 0, 0, 0.7], ...
'EdgeColor', [1, 0, 0, 0.7]);
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jon erickson
jon erickson 2024-5-15
4th color argument -brilliant solution for those of us with older versions. Thank you!

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Jan
Jan 2019-1-16
编辑：Jan 2019-1-16
A rectangle object does not have XData and YData properties and it does not allow to set the transparency by Alpha blending. This means clearly, that you cannot use rectangle to solve your problem, although it looks almost like you want it.
All you need is to define the XData and YData according to your inputs. The equations for the rounded corners can be obtained by some simple equations. I'm not sure, how the "Curvature" of rectangle objects is defined. It should be easy to find the required circles, but it is a tedius work.
Chris L'Esperance
Chris L'Esperance 2019-2-17
Ran in:
This seems like a relatively efficient way to build the polygon which can be plotted as a line or patch. If this is unneccesarily complex, please feel free to point out.
figure;
axis equal
hold on
X = [300, 600];
Y = [100, 200];
radius = 0.05;
edge_color = 'k';
% compute the dx
dx = X(1,2) - X(1,1);
% compute the dy
dy = Y(1,2) - Y(1,1);
% reduce X and Y by radius
X_reduced = [X(1,1) + (radius .* dx), ...
X(1,2) - (radius .* dx)];
Y_reduced = [Y(1,1) + (radius .* dx), ...
Y(1,2) - (radius .* dx)];
d_theta = pi/50;
theta = 0:d_theta:pi/2;
% initialize the complete series
x_rect = [];
y_rect = [];
% for each corner solve the circle equation
for vertex=1:4
% start at top right vertex, the arc that we want to first will run
% from 0 to pi/2 radians
% we cycle through the vertices in a counter-clockwise sense so that
% as we increment theta by pi/2, we arrive at the arc corresponding
% to the rounded corner
if (vertex == 1)
% top right
x = max(X_reduced); y = max(Y_reduced);
elseif (vertex == 2)
% top left
x = min(X_reduced); y = max(Y_reduced);
elseif (vertex == 3)
% bottom left
x = min(X_reduced); y = min(Y_reduced);
elseif (vertex == 4)
% bottom right
x = max(X_reduced); y = min(Y_reduced);
end
% plot circle arc
xunit = (radius .* dx) * cos(theta) + x;
yunit = (radius .* dx) * sin(theta) + y;
% add current sector to series
x_rect = cat(2, x_rect, xunit);
y_rect = cat(2, y_rect, yunit);
% increment theta
theta = theta + pi/2;
end
% close the polygon
x_rect = cat(2, x_rect, x_rect(1,1));
y_rect = cat(2, y_rect, y_rect(1,1));
plot(x_rect, y_rect, ...
'Color', edge_color);
Walter Roberson
Walter Roberson 2019-2-17
H W
H W 2023-6-5
IN YOU FIGURE, you select the input rectange, open Attribute checker, set the facealpha. ok!

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