How to change the codes to adapt to the change of graphic handle changes in r2014b?

The following codes work well in r2014a. But it does not work in r2014b because the graphic handle has been changed. So would there be a quick way to change the codes? Any help would be highly appreciated.

[c,h]=contour3(peaks(30));
for i=1:length(h)
    zz=get(h(i),'Zdata');
    set(h(i), 'ZData', 0* ones(size(zz)));
end
view(3);

Answer 1

[Walter Roberson]

The replacement code for that is

contour(peaks(30));
view(3);
grid on

In the few releases before R2014b, contour() created a contourgroup and contour3() created patch objects (in older releases yet contour() produced patch as well). However, as long as you are not needing to distinguish between contour3() and contour() on the basis that the pre-R2014b contour3() cannot have filled contours turned on easily the way contour() can, or otherwise needing to mangle the patch properties, then contour() is effectively the same as contour3() except that contour() already has the z set at 0.

If you had happened to want to set the z to something other than 0, then I have no idea how to implement that from R2014b onward.

[Mike Garrity]

Probably the simplest way to place a 2D contour at something other than Z==0 in R2014b or later would be to use hgtransform.

h1 = hgtransform('Matrix',makehgtform('translate',[0 0 5]))
[~,h2] = contour(peaks,'Parent',h1)
zlim([0 5])
view(3)

[Walter Roberson]

Good point, Mike.

[c cl]

Thanks for your kind answer! I need to do some scientific visualization of a matrix data, including two surfaces(one is 2D and the other is 3D) and two contours(one is 2D and the other is 3D). The 2D contours need to be settled on the specific z level(0 and others such as -20). The 3D contours also need to change position according to the ratio(such as 0.8 and 1.5). This code works well in r2014a. Now I want to use r2015b but this does not work. So I still use r2014a now to draw the graphic. Thanks again!

[c cl]

I just find that the 2D contour is easy to be settled in R2015b using 'ContourZLevel' as the following codes:

clc;clear;
[c,h]=contour(peaks(30));
h.ContourZLevel=-20;
view(3);
box off;
grid on;

The result is shown below:

But the 3D contour is really hard to deal with... Thanks for your help!

[Walter Roberson]

It might be easiest for you to show the code you have for the 3D contour . Or is that what you posted already? If it is then what about the solution of using contour() does not work for you?

[c cl]

Thanks again!Here is the code for 3D contour:

clc;clear;
figure;
[C3,H3]=contour3(peaks(30),10);
ra=1.2;
for n3=1:length(H3)
    hh=get(H3(n3),'Zdata');
    newhh=hh*ra;
    set(H3(n3),'Zdata',newhh,...,
        'edgecolor',[0 0 0],'linewidth',1.2);
end
grid off;

I can change the position of each contour line according to the "ra". Besides, it is also possible to change the color of each line as I want. Now I still use it in R2014a and the result is very good!

[Walter Roberson]

Refer back to the approach Mike Garrity showed above using hgtransform.