It is not clear from your comment what command you’re trying to use. You basically have 4 alternatives:

1. To use the builtin saveas function
2. To use the builtin exportgraphics function (Matlab R2020a or newer only)
3. To use the export_fig utility
4. To use the screencapture utility

First of all let me thank you for your posts they were very helpful.
I implemented a code to show contour labels with scientific notation in latex format “mantissa\cdot 10^{exponent}” based on your guidelines. It was very tedious as matlab resets these labels with many things (even not doing it when debugging the function for the plot but yes when invoking it).
Currently I reach the figures with contours seen with the abovementioned format. I can save them manually clicking in saveas but… when I try to save the figure automatically from a script using command , the figure in this case updates to the previous contour labels format.
Unbelievable! Did you find any solution for this?
Thanks!

@Shan – The ax.YDir='reverse' command, and any other similar command that repaints the axes, reverts the EdgePrims to their default (documented) behavior. So after any such command, you need to redo the setting of the undocumented functionality (set(hContour.EdgePrims...)

[X,Y,Z] = peaks;
[C,hContour] = contour(peaks,[-8 -6 -4 -2 -0 2 4 6 8], 'ShowText','on'); drawnow
set(hContour.EdgePrims(hContour.LevelList<0),'LineStyle','dotted');
ax=gca;
ax.YDir = 'reverse';

So do you have any solution for this?
Thanks

@Alon
This method runs very slow for a complex contour (e.g. not necessarily in size but perhaps one generated from real, noisy, data), which led me to experiment with some other methods. In the case where contour levels are specified to the function, I found a heuristic that seems to work:

[c,hContour] = contourf(imgaussfilt(noisyData,2),contourLevels);
hL = hContour.EdgePrims;
whichLevels = sum(bsxfun(@eq,c(1,:)',contourLevels)) > 10;

>> length(contourLevels)
ans =
    32

>> length(hL)
ans =
    24

>> length(contourLevels(whichLevels))
ans =
    24

I’m sure this might fail on some other contour plots, but it has worked on several I generated. Thoughts?

k = 1;
activecontours = [];
while k < size(hContour.ContourMatrix,2)
    activecontours = [activecontours hContour.ContourMatrix(1,k)];
    k = k + hContour.ContourMatrix(2,k) + 1;
end
levels = unique(activecontours);

Then use the levels generated in the code you supplied.

Best regards,
Alon

Unlike contour, the contourf function returns a list of levels that includes the minimum value, which does not correspond to any contour line ([-6.5466,-6,-5,-4,...] rather than [-6,-5,-4,...]). So the solution for contourf is simply to equate levels(2:end) with lines, as follows:

[c,hContour] = contourf(peaks, 'LevelStep',1, 'ShowText','on'); drawnow
levels = hContour.LevelList(2:end);
lines  = hContour.EdgePrims;
set(lines(levels<0),  'LineStyle', 'dotted');
set(lines(levels==0), 'LineWidth', 2);

Hi Yair:

Many thanks for your helpful answer. I found a problem though when using contourf.

[c,h] = contourf(peaks,'LevelStep',1); drawnow
levels = h.LevelList;
lines  = h.EdgePrims;
set(lines(levels<0),  'LineStyle', 'dotted');
set(lines(levels==0), 'LineWidth', 2);

Instead of highlighting the zero line it is the "1" line the one that appears wider.
It might be related to the different size of the levels and lines arrays. For example if I tried to change the linestyle using:

[c,h] = contourf(peaks,[-9:1:9]);
levels = h.LevelList;
lines  = h.EdgePrims;
set(lines(levels<0), 'LineStyle', 'dotted');

there are dotted lines with positive values.

Regards,
Elbio.

@Elbio – in HG2 (R2014b or newer) the contour lines (hContour.EdgePrims) correspond to the contour levels (hContour.LevelList). For example, to make all negative contour lines dotted, you can do the following:

[C,hContour] = contour(peaks, 'ShowText','on', 'LevelStep',1); drawnow
set(hContour.EdgePrims(hContour.LevelList<0), 'LineStyle', 'dotted');

I added an explanation and screenshot to the main article text above.