# how to calculate mean of interrupted data

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osasunmwen efosa am 11 Mai 2023
how can one program matlab to calculate the zero mean of a time series but only for values before a NaN value and then values after a NaN value. i am not talking about the omitnan function.
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Dyuman Joshi am 12 Mai 2023
I am not sure if I understand what you want to achieve.
Let's assume this to be your data -
A=1:20;
A([2 4 8 16]) = NaN
A = 1×20
1 NaN 3 NaN 5 6 7 NaN 9 10 11 12 13 14 15 NaN 17 18 19 20
What should be the output for this?
osasunmwen efosa am 12 Mai 2023
Bearbeitet: osasunmwen efosa am 12 Mai 2023
lets assume data A is a tracking data and the NaNs are points in time when there were interruptions. what i am trying to achieve is, the zero mean of the period whe tracking was not interrupted.
The idea is, the appearance of a NaN value marks the end of a tracking period when it comes after a real value, and if a real value comes after the NaN, that signifies the start of another tracking period.
I need to get the zero mean of these indivtracking periods i.e the mean of the periods minus the individual real values that make up that period. And my end result should be a 1x20 data like A where there will be NaNs in positions corresponding to that of A
Is this clear now ?

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### Akzeptierte Antwort

Dyuman Joshi am 12 Mai 2023
A=1:20;
A([2 4 8 16]) = NaN;
disp(A)
1 NaN 3 NaN 5 6 7 NaN 9 10 11 12 13 14 15 NaN 17 18 19 20
idx = [0 find(isnan(A)) numel(A)+1]
idx = 1×6
0 2 4 8 16 21
for k = 1:numel(idx)-1
%Range of indices between starting point, NaNs and ending point
arr=idx(k)+1:idx(k+1)-1;
A(arr) = A(arr) - mean(A(arr));
end
disp(A)
Columns 1 through 19 0 NaN 0 NaN -1.0000 0 1.0000 NaN -3.0000 -2.0000 -1.0000 0 1.0000 2.0000 3.0000 NaN -1.5000 -0.5000 0.5000 Column 20 1.5000
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Dyuman Joshi am 3 Jun. 2023
"it turns out that the standard deviation of the result is different from the standard deviation of the original dataset."
Yes, that is expected as we are manipulating the data.
"Is there a way to preserve the standard deviation of the orignal data ?"
You can preserve the standard deviation of the original data, but that will result in a different output.
osasunmwen efosa am 3 Jun. 2023
ok. please how do i preserve the standard deviation then.

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### Weitere Antworten (1)

Antoni Garcia-Herreros am 11 Mai 2023
Hello,
You could try something like this:
A=[1:10]; % For the example
A(5)=NaN; A(8)=NaN
A = 1×10
1 2 3 4 NaN 6 7 NaN 9 10
R=[1 find(isnan(A)) length(A)]; % Indices where the nans are + the beggining and end of vector
MeanVec=zeros(size(R,2)-1,1); % Initialize vector where the means will be stored
for i=1:length(R)-1 % Loop through the different sections between nans
F(i)=nanmean(A(R(i):R(i+1)));
end
F
F = 1×3
2.5000 6.5000 9.5000
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osasunmwen efosa am 12 Mai 2023
thank you for the suggestion. My dataset has many consecutive NaNs and this process only gives the mean and not the zero mean. The mean it gives still has NaNs values

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