n=100*bsxfun(@rdivide,[histc(data1,1:3) histc(data2,1:3) histc(data3,1:3)], ...
[length(data1) length(data2) length(data3)]).';
hBarH=bar(n,'hist');
xpts=cell2mat(get(hBarH,'XData'));
xpts=mean(reshape(xpts(1:2:end),2,[]))
set(gca,'xtick',xpts,'xticklabel',[1:3])
ylim([0 50])
y=n.'; y=y(:);
hTxt=text(xpts,y(:),num2str(y(:),'%.1f%%'), ...
'horizontalalign','center', ...
'verticalalign','bottom', ...
'fontsize',8);
to get
NB: See the more extensive discussion in Comment about the logic behind the above positioning and caveats re: HG vis a vis HG2 about what you can/cannot possibly be able to do with the latter.
NB2: The default 'group' option, while I can't tell the result apart from 'hist' visually, results in bar returning an array of hggroup objects, the patch objects of which are children. This makes retrieving their 'XData' property another level of indirection; hence I force the 'hist' style to get the vector of patch objects directly instead.
ADDENDUM
I just noticed the trees for the forest--have always concentrated on how to retrieve the necessary data so didn't pick up on what looks to be a simple algorithm to compute the bar positions. It looks from this sample of one that the delta to the various bars from the center x position is (N-1)/(M*N) where N is number of bars/group and M is number of groups.
That would be readily computed without handle-diving if it holds in general--
>> dx=(size(n,1)-1)/numel(n);
>> xpts=[[1:3].'-dx [1:3].' [1:3].'+dx].'; xpts=xpts(:).'
xpts =
0.7778 1.0000 1.2222 1.7778 2.0000 2.2222 2.7778 3.0000 3.2222
>>
Those are the values we obtained from the averages of the patches x-axis locations above so looks like we've uncovered the magic Rosetta stone used by the internal logic. Try this out with HG2; if it works as I suspect it will, then the problem about opaque objects at least has a relatively simple workaround for this particular issue.
A NOTE: I tried some other sizes of groups and bars/group and while the above gets in the ballpark, the spacings aren't that simply derived for them so the above is true only for the case of the 3x3 arrangement it appears, unfortunately.
