Vector averaging with mean command

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Lev Mihailov
Lev Mihailov 2020-2-3
评论： Lev Mihailov 2020-2-3
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
  2 个评论
KSSV
KSSV 2020-2-3
what do you mean by averaging over 10, 50 and 100?
Lev Mihailov
Lev Mihailov 2020-2-3
average vector by blocks of 10 and 50
X=[1:100]
Averaging1=mean(X,10);
Averaging1=[5.5 15.5 ....]

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回答（2 个）

Rik
Rik 2020-2-3
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)
  3 个评论
显示 1更早的评论
Rik
Rik 2020-2-3
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))]
Lev Mihailov
Lev Mihailov 2020-2-3
I got a vector of values ​​(necessary areas) that need to be averaged

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Bhaskar R
Bhaskar R 2020-2-3
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));
  2 个评论
Lev Mihailov
Lev Mihailov 2020-2-3
编辑：Lev Mihailov 2020-2-3
X=[1:1:553];
Averaging1=mean(X(1:10));
such a solution is good, but can you use a loop to make a vector along the entire length of the vector?
Averaging1=mean(X(1:10)); Averaging1=mean(X(11:21)); Averaging1=mean(X(22:32));
Rik
Rik 2020-2-3
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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