Mean value of data with irregular intervals
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Hello,
I am trying to obtain average y value for a data set (x,y). However, interval between data points is not regular and mean(y) is off from the expected value. For example,
x = [1 1.1 1.2 1.3 2 3 4 5.1 5.2 5.3 5.4]; y = [0 0.1 0.2 0.3 1 2 1 0.3 0.2 0.1 0];
Because x,y are more densely lying in x = [0 1] and [5 6] intervals, y values in those intervals are weighted more if I take mean(y). Do you know how to avoid this and obtain y value in uniform intervals?
Thanks, Bobo
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採用された回答
Iain
2013 年 7 月 29 日
Why not just take a weighted mean?
This method ignores y(1) - you could change it to ignore y(end), or add info and inclue it.
totalweight = 0;
weightedsum = 0;
for i = 2:numel(x)
weight = x(i)-x(i-1);
weightedsum = weightedsum + weight*y(i);
totalweight = totalweight + weight;
end
meanvalue = weightedsum / totalweight;
その他の回答 (1 件)
dpb
2013 年 7 月 29 日
編集済み: dpb
2013 年 7 月 29 日
One way to weight uniformly...
dx=min(diff(x)); % shortest difference in vector
xi=x(1):dx:x(end); % fill in all values equally
yhat=interp1(x',y',x1'); % and get corresponding y's...
mean(yhat); % obviously can combine steps ...
2 件のコメント
Andrew Sandeman
2023 年 6 月 15 日
In my opinion this is the correct answer, not the currently accepted answer.
- This method does not neglect y(1)
- Samples which are obtained more sparsely are weighted more accurately using this interpolation method
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