mean value of array
显示 更早的评论
I would appreciated if someone can tell me what am I doing wrong or maybe ?
all the matrix are 20x1 in my varibles
[Relative_ErrorM, indexAtM] = mean(Relative_Error,'omitnan');
r_experM= r_exper(indexAtM,);
plot(r_experM,Relative_ErrorM, 'kv', 'LineWidth', 2, 'MarkerSize', 15);
text(r_experM,Relative_ErrorM,sprintf(' Mean = %.1f', Relative_ErrorM))
here is the error message
Error using mean
Too many output arguments.
采纳的回答
更多回答(1 个)
Image Analyst
2019-12-15
0 个投票
Note that the mean of an array won't necessarily fall at ANY index. It's quite possible the mean is not one of hte numbers in the array. For example mean([1,2]) is 1.5 but 1.5 is not in the array. Did you perhaps mean min() or max() instead of mean()? Those functions can return the index where the first occurrence of the min or max appears.
2 个评论
amjad almaarafawi
2019-12-15
Image Analyst
2019-12-16
No, since you don't have an "x" value. However, you can put a marker on the actual value that is closest to the mean.
plot(x, y, 'b-', 'LineWidth', 2); % Plot curve.
% Find mean y value.
meanYValue = mean(y);
% Find out differences between actual y values and the mean y value.
diffs = abs(y - meanYValue);
% Find the index the y value is closest to the mean y value.
[minDiff, indexAtMinDiff] = min(diffs);
% Plot a circle around that point.
hold on
plot(x(indexAtMinDiff), y(indexAtMinDiff), 'ro', 'MarkerSize', 20);
类别
在 帮助中心 和 File Exchange 中查找有关 Creating and Concatenating Matrices 的更多信息
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!