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How to extract x,y,z coordinates from a contour?

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Jay
Jay on 19 Apr 2017
Edited: Adam Danz on 11 Feb 2020 at 20:52
Hi everybody,
I used the following codes to extract X,Y,Z coordinates from a contour. When I ran it in MATLAB 2013a, it works. But when I ran it in MATLAB 2016a, it doesn't work. I found that the codes is very sensitive to MATLAB version. Can anybody tell me how to solve this problem? Or are there any other MATLAB codes that serve the similar function? The code is from this link:

  2 Comments

Adam Danz
Adam Danz on 23 Jan 2020
In addition to the answers below, see this file exchange function that returns a table of (x,y) coordinates of the contour lines.

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Accepted Answer

Star Strider
Star Strider on 19 Apr 2017
This works in R2017a, so it should also work for R2016a.
Example
f = @(x,y) x.^2 - y.^2; % Your Function
x = linspace(-10, 10); % X-Range
y = linspace(-10, 10); % Y-Range
[X,Y] = meshgrid(x,y); % Create Matrix Arguments
figure(1)
[C,h] = contour(X, Y, f(X,Y));
grid
Lvls = h.LevelList;
for k1 = 1:length(Lvls)
idxc{k1} = find(C(1,:) == Lvls(k1));
Llen{k1} = C(2,idxc{k1});
conturc{k1,1} = C(:,idxc{k1}(1)+1 : idxc{k1}(1)+1+Llen{k1}(1)-1);
conturc{k1,2} = C(:,idxc{k1}(2)+1 : idxc{k1}(2)+1+Llen{k1}(2)-1);
end
figure(2)
plot(conturc{3,1}(1,:), conturc{3,1}(2,:))
grid
The contour data now are a (2xN) vector created by horizontally concatenating the (x,y) coordinates of each contour line. The segments are created each as:
C = [Level x-coordinates; Length y-coordinates]
so the first column of each segment are the ‘Level’ and the ‘Length’ of the segment. The second output are the properties of the plot.
My code first plots the contour function, returning both outputs. It then uses the ‘LevelList’ property to return the plotted levels, and then uses these in the loop to find the start indices of the segments in the ‘C’ matrix in the ‘idxc’ cell array, and the number of elements in each contour segment in the ‘Llen’ cell array. It then uses these to extract the x (first row) and y (second row) coordinates for each contour segment in the ‘conturc’ cell array. The z-coordinates are the levels. They do not appear in the cell arrays, but correspond to the values in the ‘Lvls’ vector.
The plot in figure(2) is not necessary for the code. It shows how to get the data from the cell arrays (in this instance the third cell in the first group), and that the extracted values are correct.
This turned out to be an interesting adventure!

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Star Strider
Star Strider on 19 Apr 2017
My pleasure.
As of R2014b, graphics in MATLAB changed from the original handle graphics, known now as HG1, to a second-generation version with significant changes, known as HG2. The transition was significant, and ‘broke’ a lot of pre-existing code in many graphics functions on the File Exchange and in MATLAB Answers.
There are no other functions that I am aware of that can extract the data from the R2014b and later contour function, although I did not search the File Exchange. I wrote this code specifically to Answer your Question., and is original to me and this Answer.
If my Answer solved your problem, please Accept it!
——————
EDIT
Additionally, if you want to include the ‘Lvls’ variable with each cell array element, it is easy to add it:
for k1 = 1:length(Lvls)
idxc{k1} = find(C(1,:) == Lvls(k1));
Llen{k1} = C(2,idxc{k1});
conturc{k1,1} = C(:,idxc{k1}(1)+1 : idxc{k1}(1)+1+Llen{k1}(1)-1);
conturc{k1,2} = C(:,idxc{k1}(2)+1 : idxc{k1}(2)+1+Llen{k1}(2)-1);
conturc{k1,3} = Lvls(k1);
end
To use it with the rest of your data, address it as you would the other data:
figure(2)
plot(conturc{3,1}(1,:), conturc{3,1}(2,:), '-b')
hold on
plot(conturc{3,2}(1,:), conturc{3,2}(2,:), '-b')
hold off
legend(sprintf('Level = %.1f', conturc{3,3}), 'Location','E')
grid
This uses the legend function, and can be adapted to the text function if you want to.

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More Answers (1)

Rik
Rik on 19 Apr 2017
The culprit is in this code snippet:
[c,h]=contour(X,Y,Z,v);
xcg=get(get(h,'children'),'xdata');
ycg=get(get(h,'children'),'ydata');
The form in which h is returned is slightly different between 2013a and 2016a. The first returns a handle, the second an object. Usually this doesn't matter, but it does in this case.
I compared the results from [c,h]=countour(flipud(P)) between 2012b and 2017a (after load penny;). Apparently, the newer object no longer has a child for each contour element, so you will have to put in some effort to add in the NaNs yourself to get back the original behavior.

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Rik
Rik on 19 Apr 2017
Star Strider did most the heavy lifting for you.
The conturc cell array is almost equal to the xcg and ycg variables that you can find in that FEX submission. You only need to add a NaN to the end of each matrix and split conturc. In short, replace these lines
[c,h]=contour(X,Y,Z,v);
xcg=get(get(h,'children'),'xdata');
ycg=get(get(h,'children'),'ydata');
with these lines
[c,h]=contour(X,Y,Z,v);
%get the Matlab version, add 0.5 to the year for the b release
MatlabVersion=version('-release');
MatlabVersion=str2double(MatlabVersion(1:(end-1)))...
+0.5*strcmp(MatlabVersion(end),'b');
if MatlabVersion<=2014%2014a is somewhere around the switch in behavior
xcg=get(get(h,'children'),'xdata');
ycg=get(get(h,'children'),'ydata');
else
Lvls = h.LevelList;
conturc=cell(length(Lvls),2);%pre-allocate variable
for k1 = 1:length(Lvls)
idxc{k1} = find(c(1,:) == Lvls(k1));
Llen{k1} = c(2,idxc{k1});
conturc{k1,1} = C(:,idxc{k1}(1)+1 : idxc{k1}(1)+1+Llen{k1}(1)-1);
conturc{k1,2} = C(:,idxc{k1}(2)+1 : idxc{k1}(2)+1+Llen{k1}(2)-1);
end
conturc=conturc(:);
for n=1:length(conturc)
conturc{n}=[conturc{n} NaN(2,1)];%add a NaN column
end
xcg=conturc;%this is a copy, containing both x and y
ycg=conturc;%this is a copy, containing both x and y
for n=1:length(xcg)
xcg{n}=xcg{n}(1,:)';%remove y
ycg{n}=ycg{n}(2,:)';%remove x
end
end

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