Hey,
I have a vector containing the mean values and a another vector with the standard deviations. I want to plot the standard deviation as a shaded area and the mean as a line as shown on the image below but I want to write my own function. hope someone can help
stdshade.PNG

2 件のコメント

M
M 2019 年 12 月 3 日
I want to write my own function
Does this mean you do not want to use built-in function like fill or area ?
You can start by reading this discussion:
https://mathworks.com/matlabcentral/answers/180829-shade-area-between-graphs
Hinna Ahmed
Hinna Ahmed 2019 年 12 月 3 日
I want to use built-in functions like fill or area but not functions like stdshade because they use a matrix with all the observations as a input but i have already calculated the mean and std.

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M
M 2019 年 12 月 3 日
編集済み: M 2019 年 12 月 3 日

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You can use the code directly from this answer:
y = rand(1,10); % your mean vector;
x = 1:numel(y);
std_dev = 1;
curve1 = y + std_dev;
curve2 = y - std_dev;
x2 = [x, fliplr(x)];
inBetween = [curve1, fliplr(curve2)];
fill(x2, inBetween, 'g');
hold on;
plot(x, y, 'r', 'LineWidth', 2);

4 件のコメント

Baldvin
Baldvin 2025 年 11 月 14 日
Here's the a similar code as above, using Patch and its properties. Hopefully, the variable names are descriptive:
% Dummy data:
x = linspace(-2,2,64);
meanCurve = 2 + x.^2 + 0.55*rand(size(x));
upperCurve = meanCurve + 0.5*abs(randn(size(x))) + 1;
lowerCurve = meanCurve - 0.5*abs(randn(size(x))) - 1;
% Create shaded area. We just have to create x/y points that trace out the
% bottom curve from left to right, followed by the upper curve from right to left:
areaX = [x, fliplr(x)];
areaY = [lowerCurve, fliplr(upperCurve)];
shadedArea = patch("XData",areaX, "YData",areaY, ...
"FaceColor",'flat',"EdgeColor",'flat',"SeriesIndex",2, ...
"FaceAlpha",0.2, "EdgeAlpha",1);
% Add the curve:
hold on
meanLine = plot(x, meanCurve, "SeriesIndex",2, "LineWidth", 2);
Star Strider
Star Strider 2025 年 11 月 14 日
This works for row vectors:
areaX = [x, flip(x)];
areaY = [lowerCurve, flip(upperCurve)];
For column vectors, use:
areaX = [x; flip(x)];
areaY = [lowerCurve; flip(upperCurve)];
NOTE -- The flip function will flip a vector appropriately. Use flipud or fliplr if you are working with matrices. (Using fliplr on a column vector does nothing useful!)
.
Paige
Paige 2026 年 9 月 16 日 17:15
編集済み: dpb 2026 年 9 月 16 日 17:20
I am having similar problems with my code and I think its because my x axis is not numerical, rather a date and time.
This is the plot that was produced from the following code:
peakdir = data(:,8);
peakdirspr = data(:,9);
upperp = peakdir + peakdirspr/2;
lowerp = peakdir - peakdirspr/2;
areaX = [time, fliplr(time)];
areaY = [lowerp, fliplr(upperp)];
shadedArea = patch("XData",areaX, "YData",areaY, ...
"FaceColor",'flat',"EdgeColor",'flat',"SeriesIndex",2, ...
"FaceAlpha",0.2, "EdgeAlpha",1);
hold on
meanLine = plot(time, peakdir, "SeriesIndex",2, "LineWidth", 2);
This was code that I got from another thread, but I was exactly able to make it work. Any input would be greatly appreciated!
dpb
dpb 2026 年 9 月 16 日 17:29
",,,I think its because my x axis is not numerical, rather a date and time."
No, the problem is that the last y data being lower in magnitude when flipped back to the origin to create the boundaries for the patch object is a straight line between the two points. The "jaggies" in the time trace are the underlying problem.
It would be helpful to attach the data file so folks have something to work with instead of having to make up simulated data to illustrate potential solutions.

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