How to determine average and standard deviation of y axis values for corresponding bin of x axis?
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trailokya
el 16 de Sept. de 2015
Comentada: Image Analyst
el 22 de Sept. de 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 comentarios
Star Strider
el 16 de Sept. de 2015
I explained this in my Answer to your previous post (that you have as yet to Accept).
Kelly Kearney
el 16 de Sept. de 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');
Respuesta aceptada
Image Analyst
el 16 de Sept. de 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 comentarios
Image Analyst
el 22 de Sept. de 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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