Computation of mean and standard deviation after supressing NaNs of an array

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nnmax = 4;
for nn=2:nnmax
D_filt(nn,:) = D_filt(nn-1,:);
[Dmax(nn,aa),Imax(nn,aa)] = max(D_filt(nn-1,:));
D_filt(nn,Imax(nn,aa)) = NaN ;
end
#Comments (loop indicates)
1. Scans from 0 to 360 degrees with 10 degree interval.
2. aa represents the same [0:10:360]
3. At each value of angle, the detector collects 10 data points
4. OBJECTIVE: To discard 4 data points out of the 10 and compute mean and SD
5. nn represnts the iteration number while supressing the bad data points
6. The output of the above loop displays
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 0 0
1 2 3 4 5 6 7 0 0 0
Instead of 4 rows, a single row matrix after supressing four bad points is desired, so that mean and SD can be computed with ease.

Answers (2)

Thomas Koelen
Thomas Koelen on 8 Apr 2015
Edited: Thomas Koelen on 8 Apr 2015
Hi Naga,
The term "average" usually encompasses several ways to measure what value best represents a sample. There are various measurements that are used and at times the term and measurement used depends on the situation.
A statistician or mathematician would use the terms mean and average to refer to the sum of all values divided by the total number of values, what you have called the average. This especially true if you have a list of numbers. In fact even in mathematics there are different "averages" or "means" and this one is more properly called the arithmetic mean.
I think the thing you are trying to accomplish is straight up using the function:
mean()
For the standard deviation use this function:
std()
Thomas Koelen
  3 Comments
Naga
Naga on 8 Apr 2015
Edited: Naga on 8 Apr 2015
When I run the program, the output is zeroes instead of NaN, so when I use your above function, there's no change in the output.
Just to give you an idea, When I run the above loop the following is being displayed in the output
D_filt
1 2 3 4 5 6 7 8 9 10
1 2 3 4 5 6 7 8 9 0
1 2 3 4 5 6 7 8 0 0
1 2 3 4 5 6 7 0 0 0
So, there are no NaNs in principle

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Greg Dionne
Greg Dionne on 8 Apr 2015
Edited: Greg Dionne on 8 Apr 2015
You can use mean(x,'omitnan') in R2015a. If you have the Statistics Toolbox, you can use nanmean.
Otherwise, try this:
n = sum(~isnan(x));
x(isnan(x)) = 0;
s = sum(x);
avg = s./n;
  7 Comments
Greg Dionne
Greg Dionne on 9 Apr 2015
In that case you could do:
sum(x(4,:),2) ./ sum(0~=x(4,:),2)
But it seems you are still confused as to why you are not getting NaN in your output like you are expecting. Perhaps if you posted your data (save it to a MAT file and post it to your question) and a quick blurb about what you want to see happen (i.e. what you expect the answer to be) we'll be able to point you in the right direction.
Good luck!

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