Thanks, Ben. I didn't know the setappdata/getappdata pair, so I actually wrote one myself as a function, just adding my var to myhandles and attach to the figure handle. It worked fine for me so far :).
Ben, Thanks for the reply. I was trying to do this:
hf=figure;
bp = uiextras.BoxPanel('Parent',hf,...
'Tag','boxpanel');
bp.Title = 'Original Title';
hobj=findobj(hf,'Tag','boxpanel');
hobj.Title='New Title';
The objective is to be able to modify the title in a callback. I thought I could use the handle returned by findobj with tag string, but that handle seems not working. I think it due to the fact that the handle returned by findobj is the handle to the container of the composite GUI object, so it's not directly related to the title banner. Now I use guidata to save the actual handle of the boxpanel, then retrieve it inside the callback, and it works fine.
I don't know how difficult it is to modify the layout toolbox source code to make this happen, I will spend sometime look it up after finishing current task.
Thanks again for sharing a great GUI tool.
One of the best!
I have one question related to the BoxPanel, I was trying to modify the 'Title' through hobj.Title='new Title', but it doesn't work. Is there a way to accomplish this? Thanks!
John,
Welcome to the fun of floating-point arithmetic. First, the duplicate points *are* removed, but this is not obvious unless you display all the digits of the result. For your third example, the four "duplicates" are actually all different, but only in the final bit (you can see it with "format hex"). I don't think there's anything I can do to make this work correctly in all cases.
Second, why should something fail if an intersection occurs at a segment boundary? Some people would like to consider this a proper intersection and others not (I know this from several discussions on the topic). As a user, you will have to evaluate the results you get for your particular application.
In many real world applications, where the two curves are formed from real data, this issue never comes up at all. It's (mostly) only from textbook type problems that you'll see this behavior.
Doug
Douglas,
Thanks for the nice function.
Could you please clarify the behavior depending on the ROBUST flag?
From the function doc:
% ROBUST (optional) set to 1 or true means to use a slight variation of the
% algorithm that might return duplicates of some intersection points, and
% then remove those duplicates. The default is true, but since the
% algorithm is slightly slower you can set it to false if you know that
% your curves don't intersect at any segment boundaries. Also, the robust
% version properly handles parallel and overlapping segments.
First, based on the first sentence, it would seem that the removal of duplicates is handled by intersections.m. This is not the case.
Second, it would seem that something should fail if the curves *do* intersect at segment boundaries (points specifying the line) and the ROBUST = 0/false is used.
At least for the example below, this is not the case.
Thanks for your help!
% Test intersections function
% Depending on whether ROBUST is true or false, and depending on how points
% defining each line do or don't lay (exactly = machine prec?) on the other
% line, the algorithm returns a different number of intersections.
clear all
% % two lines where no point is exactly on other line
% x1 = [0.1, 0.4, 0.5, 0.8];
% y1 = [0.2, 0.5, 0.7, 0.9];
% x2 = [0.1, 0.4, 0.8];
% y2 = [0.9, 0.3, 0.1];
% % first line has point on the other, but not vice versa
% x1 = [0.1, 0.25, 0.5, 0.8];
% y1 = [0.2, 0.6, 0.7, 0.9];
% x2 = [0.1, 0.4, 0.8];
% y2 = [0.9, 0.3, 0.1];
% % two lines that share a common point
% x1 = [0.1, 0.4, 0.5, 0.8];
% y1 = [0.2, 0.3, 0.7, 0.9];
% x2 = [0.1, 0.4, 0.8];
% y2 = [0.9, 0.3, 0.1];
robust = 0;
[xinter, yinter, iinter, jinter] = intersections(x1, y1, x2, y2, robust);
figure
hold on
scatter(xinter, yinter, 200, 'r','filled')
plot(x1, y1, '.b-','markersize',30)
plot(x2, y2, '.g-','markersize',20)
text(0.5, 0.5, ['found ', num2str(numel(xinter)), ' intersection/s'], 'fontsize',14)
did anyone used this code just for one curve? could you help me with the transformation? i am new in matlab, and i can't achieve a working code for just one curve. thank you so so much.
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