# How to determine the x-axis positions of the bars in a grouped bar chart

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the cyclist on 16 Apr 2015
Commented: Alex Wooten on 16 Mar 2021
Suppose I make the following bar chart:
figure; bar([1 2],[3 4; 5 6])
How can I programatically find the x-axis locations of each of the bars? I don't want x = [1 2]. I want the actual locations of the blue and yellow bars.

Allison on 3 Jun 2015
the cyclist-
I ran into this same problem when switching from Matlab 2011b to 2015b, with the updated graphics engine. After fooling around for a bit, I found that the following code will work (adapted from your original post):
%
mean_velocity = [5 6 7; 8 9 10]; % mean velocity
std_velocity = randn(2,3); % standard deviation of velocity
figure
hold on
hb = bar(1:3,mean_velocity');
% For each set of bars, find the centers of the bars, and write error bars
pause(0.1); %pause allows the figure to be created
for ib = 1:numel(hb)
%XData property is the tick labels/group centers; XOffset is the offset
%of each distinct group
xData = hb(ib).XData+hb(ib).XOffset;
errorbar(xData,mean_velocity(ib,:),std_velocity(ib,:),'k.')
end
I'm not quite sure why the pause is required- it seems that the new graphics engine requires a little bit of time before it will recognize the figure handle properly. I've tried it in other scripts where there is more code in between the plotting and the additional math and it seems to work fine, as well as trying it just via command line, and found the same issue. This should work in any version r2014b+.
Alex Wooten on 16 Mar 2021
+1 for comment about pause. I had plotting code giving me different results when running line-by-line vs all at once, that explains it!

Maxandre Jacqueline on 27 Aug 2018
How to do this for a categorical x axis?

Hossein on 16 Nov 2017
Edited: Hossein on 16 Nov 2017
Hi, Thank you for your answer, but unfortunately, there is no 'XOffset' property for the handle in Matlab 2017, does anyone have other ideas?
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Hossein on 21 Sep 2018
Mine was 2017a, now I tested on 2017b and it worked. :) Thank you for your reply.