How to determine average and standard deviation of y axis values for corresponding bin of x axis?
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trailokya
on 16 Sep 2015
Commented: Image Analyst
on 22 Sep 2019
Sir i have two data series say
a b
25.36737061 -27.47956892
20.54391479 -23.68162398
16.76391602 -16.65254461
9.47177124 -19.20600915
16.25158691 -18.56570783
4.462646484 -14.39363913
7.785919189 -14.98048449
12.27481079 -18.49125231
4.851806641 -19.91135093
2.111236572 -5.049334665
-1.457702637 -6.51219601
1.85055542 -1.299793246
and i want to plot for bin width of 'a' (say 5,10,15 etc) with the corresponding average and standard deviation of b. Please help me with a easy way since the series are very large?
5 Comments
Star Strider
on 16 Sep 2015
I explained this in my Answer to your previous post (that you have as yet to Accept).
Kelly Kearney
on 16 Sep 2015
To simplify the tracking of indices, you may want to look at this aggregatehist.m function... it's basically a wrapper around the histc plus accumarray method demonstrated by Star Strider and myself.
ab = [...
25.36737061 -27.47956892
20.54391479 -23.68162398
16.76391602 -16.65254461
9.47177124 -19.20600915
16.25158691 -18.56570783
4.462646484 -14.39363913
7.785919189 -14.98048449
12.27481079 -18.49125231
4.851806641 -19.91135093
2.111236572 -5.049334665
-1.457702637 -6.51219601
1.85055542 -1.299793246];
bin = -5:5:30;
[aa,bb] = aggregatehist(bin, ab(:,1), ab(:,2));
dev = cellfun(@std, bb);
avg = cellfun(@mean, bb);
amid = (bin(1:end-1)+bin(2:end))./2;
plot(ab(:,1), ab(:,2), '.');
hold on;
errorbar(amid, avg, dev, 'linestyle', 'none', 'marker', 'o');
set(gca, 'xtick', bin, 'xgrid', 'on');
Accepted Answer
Image Analyst
on 16 Sep 2015
Why do you say (the badly-named) "a" is the bin width and then say it has values of 5, 10, or 15 when it clearly does not have any of those values? You can specify the edges of the bins so that the bin widths are exactly what you want. Like if you want a bin width of a(1), which is 25.3673, then you can do
edges = 0 : a(1) : 50
and then pass edges into histc() or histcounts().
5 Comments
Image Analyst
on 22 Sep 2019
Thanks Wade. Maybe so, since I seem to be immortal, what with my responses of long ago living forever. You can "vote" for answers if you want to give anybody additional "reputation points".
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