multiple area with different base values

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Is it possible to show multiple area in a graph with different base values ? In the following example, the base value of the first area is changing when the second one is displayed.
figure
x = linspace(0,2*pi);
y = sin(x);
plot(x,y,'linewidth',2);
hold on;
f = y>=0.9;
h(1) = area(x(f),y(f),...
'BaseValue',0.9,...
'ShowBaseLine','off',...
'FaceColor','red',...
'FaceAlpha',0.2,...
'LineStyle','none');
fprintf('basevalue: %.2f\n',h(1).BaseValue);
f = y<=-0.3;
h(2) = area(x(f),y(f),...
'BaseValue',-0.3,...
'ShowBaseLine','off',...
'FaceColor','green',...
'FaceAlpha',0.2,...
'LineStyle','none');
fprintf('basevalue: %.2f\n',h(1).BaseValue);
Thanks for the help ! (Matlab R2016b)
  1 个评论
Lionel
Lionel 2017-5-17
One way to bypass this issue :
figure
x = linspace(0,2*pi);
y = sin(x);
plot(x,y,'linewidth',2);
hold on;
bv = 0.9;
f = y>=bv;
ybv = repmat(bv,1,sum(f));
h = area([x(f);x(f)]',[ybv;y(f)-ybv]',...
'ShowBaseLine','off',...
'FaceColor','red',...
'FaceAlpha',0.2,...
'LineStyle','none');
delete(h(1));
f = y<=-0.3;
h2 = area(x(f),y(f),...
'BaseValue',-0.3,...
'ShowBaseLine','off',...
'FaceColor','green',...
'FaceAlpha',0.2,...
'LineStyle','none');
but...

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回答(2 个)

Manish Annappa
Manish Annappa 2017-5-19
I compared the graphs generated from the above code snippets posted by you. The second code snippet seems to give the expected results. Any reason why this solution is not acceptable to you?
  1 个评论
Lionel
Lionel 2017-5-19
Indeed the second code snipped gives the expected results but is much less intuitive. Is is a bug or a feature to have only one BaseValue per axis ?

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Ullekh Gambhira
Ullekh Gambhira 2018-6-16
I was facing a similar issue, solved it by using "Shade area between two curves" function by John Bockstege. The file exchange link for this function is below:
https://www.mathworks.com/matlabcentral/fileexchange/13188-shade-area-between-two-curves
It does not consider base value, but basically shades the area between two lines. So one line can be your vector and the other can be a baseline*ones(size(x)).
Hope this helps.
Regards, Ullekh

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