MATLAB Answers


Graph not large enough to contain thick line

さんによって質問されました 2019 年 7 月 8 日
最新アクティビティ Star Strider
さんによって コメントされました 2019 年 7 月 8 日
The grey line here shows the average +/- one standard deviation, which is why the line is so thick. I'm trying to change the axis limits of the graph such that it can contain the whole line without it overlapping with the axes, but have so far been unsuccessful. Does anyone have a good method for this? I'm not showing any code here because this is just what it looks like with matlab's automatically generated axis limits.

  6 件のコメント

2019 年 7 月 8 日
I don't know if it's simply a difference in matlab versions or something else, but in my particular case it's at least symmetrical, but you're definitely right, it doesn't reflect the actual value at the moment.
2019 年 7 月 8 日
and the line width is (hopefully) +/- one standard deviation from that data point
It's very unlikely. As documented, the LineWidth is measured in points (= 1/72 of an inch). The acutal physical width of the line compared to other objects will therefore depends on the Units property of the figure and on the Units property of the axes. Only if both resolve to points (via explicit points or derived by normalized) will the thickness matches the standard deviation.
Star's answer is probably a lot more reliable than setting the LineWidth.
Bruno Luong
2019 年 7 月 8 日
"but in my particular case it's at least symmetrical"
Same for me on screen, the dissymetry comes from the figure exportation, but it still shows my point.

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2 件の回答

Star Strider
回答者: Star Strider
2019 年 7 月 8 日

Rather than plotting a thick line, use a patch object instead for the standard deviations:
x = 0:99; % Create Data
y = sin(2*pi*x/25); % Create Data
sd = rand(size(y)); % Standard Deviations
plot(x, y)
hold on
patch([x, fliplr(x)], [y+sd, fliplr(y-sd)], [1 1 1]*0.8, 'EdgeColor','none')
hold off
If you want the ‘y’ line in top of the patch object, plot it after you plot the patch object.

  2 件のコメント

2019 年 7 月 8 日
Thank you, I didn't consider that option. I've never worked with patch objects before, so bear with me, but I'm now getting two of the same object in the same subplot.
patch([xL',fliplr(xL')],[yL+stDev fliplr(yL-stDev)],[1 1 1]*0.9,'EdgeColor','none')
Are the constraints off?
Star Strider
2019 年 7 月 8 日
My pleasure.
For my code to work correctly, ‘xL’, ‘yL’, and ‘stDev’ all must be row vectors, not column vectors, in the patch call. (They can be anything you want elsewhere in your code.) One way to be certain that they are compatible with patch is to change the patch call slightly to:
patch([xL(:)',fliplr(xL(:)')],[yL(:)'+stDev(:)', fliplr(yL(:)'-stDev(:)')],[1 1 1]*0.9,'EdgeColor','none')
I tested that patch call and it works with my test vectors. It should work in your code by simply copying it and pasting it.

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Bob Nbob
回答者: Bob Nbob
2019 年 7 月 8 日

Try here.

  1 件のコメント

2019 年 7 月 8 日
I believe the OP generated the graph by using a very large LineWidth to plot the grey line. The auto limits for axes don't take into account line thickness, so actual limits would have to be calculated manually.

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