Vector averaging with mean command
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Hello! I don’t quite understand how to use this command, I need to average a 1x533 vector
Averaging1=mean(X,10);
Averaging2=mean(X,50);
Averaging3=mean(X,100);
I need to average over 10, 50 and 100, when I use such commands nothing happens
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답변 (2개)
Rik
2020년 2월 3일
A faster method than a for-loop is using movmean and throwing out the results you don't need. This will also take care of the case where you have an incomplete trailing block.
clc
X=rand(28,1);
n=10;
tic
Averaging1=[mean(X(1:10)); mean(X(11:20)); mean(X(21:28))];
toc
tic
Averaging2=movmean(X,[0 n-1]);
Averaging2=Averaging2(1:n:end);
toc
isequal(Averaging1,Averaging2)
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Rik
2020년 2월 3일
Do you mean this should be repeated for different values of n? Or do you have a vector with all group sizes?
%example for the latter:
n=[2 4 3];
Averaging1=[mean(X(1:2));mean(X(3:6));mean(X(7:9))]
Bhaskar R
2020년 2월 3일
Second argument of mean is dimension, your input vector X is 1x533 vector 1D vector.
1) I assume you need mean of the values from 1 to 10, 1 to 50 and 1 to 100 values
Averaging1=mean(X(1:10));
Averaging2=mean(X(1:50));
Averaging3=mean(X(1:100));
or 2) I assume you need mean of the values from 1 to 10, 11 to 50 and 51 to 100 values then
Averaging1=mean(X(1:10));
Averaging2=mean(X(11:50));
Averaging3=mean(X(51:100));
댓글 수: 2
Rik
2020년 2월 3일
You are ignoring the fact that your padding influences the mean value of the final block. You should probably use something like the code below.
if mod(numel(X),n)~=0
elem_count=numel(X);
X( (end+1):(n*ceil(numel(X)/n)) )=0;
X=reshape(X,n,[]);
divs=n*ones(1,size(X,2)-1);
divs(end+1)=mod(elem_count,n);
Averaging3=sum(X,1)./divs;
Averaging3=Averaging3';
else
X=reshape(X,n,[]);
Averaging3=mean(X,1);
Averaging3=Averaging3';
end
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