Suppose I have a column vector A=[1 3 -999 4 9 8 -999 3 5]; I need to get average of each 3 sections. The problem is, if any section contains -999, I need to avoid it and get the average of rest of the numbers. For an example, for the 1st three numbers, average has to be (1+3)/2.
Please help me with that. Thanks.

3 commentaires

MSP
MSP le 28 Juil 2017
Modifié(e) : MSP le 28 Juil 2017
clear all
close all
A=[1 3 -999 4 9 8 -999 3 5]
index=find(A==-999)
A(index)=0;
sectionsize=3;
k=1;
for i=1:3:size(A,2)-2
out=A(i:i+2)
nzeros=numel(find(out==0));
output(k)=sum(out)/(sectionsize-nzeros);
k=k+1;
end
output
Thanks. But using this code, if there were three -999 somewhere in the middle, that section will be replaced by the avg of next three. At the same time, if there are three -999 at the end, it will be not shown at all. In those cases, let's put NaN so that I know which sections had three consecutive -999. Consider the following to help me with coding.
A=[1 3 -999 -999 -999 -999 4 9 8 -999 -999 -999]; The correct answer I am expecting is B=[2 NaN 7 NaN]
Thanks.
Image Analyst
Image Analyst le 28 Juil 2017
Modifié(e) : Image Analyst le 28 Juil 2017
That seems inadvisable. Why do you want nans in there instead of the very simple approach I showed you below?????
MSP, that should have been in the answer section below since you intended it as an answer rather than a comment (like asking for more information).

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Réponses (1)

Image Analyst
Image Analyst le 28 Juil 2017
If you have the Image Processing Toolbox you can use regionprops():
A=[1 3 -999 4 9 8 -999 3 5]
props=regionprops(A~=-999, A, 'MeanIntensity');
sectionMeans = [props.MeanIntensity]
You get:
sectionMeans =
2 7 4

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