hey all. i have a 5760x1 matrix. i need to find the average of the first 10 elements, remove the 10 elements and replave with the averaged number, find the average of the next 10 elements, remove those 10 elements and replace with average and so on.... could anyone give me an insight as to how to do this? much appreciated john

 채택된 답변

Sean de Wolski
Sean de Wolski 2014년 10월 15일

2 개 추천

B = mean(reshape(x,10,[]))'
Where x is your original column vector and B is the mean of every 10x1 chunk

추가 답변 (3개)

Andrew Reibold
Andrew Reibold 2014년 10월 15일
편집: Andrew Reibold 2014년 10월 15일

1 개 추천

myMatrix = rand(5760,1)
for jj = 1:576
moving_avg(jj,1) = mean(myMatrix(jj*10-9:jj*10));
end
Will make a matrix called 'moving_avg' where each point represents the average of each set of 10 points.
Is that what you wanted? It will be 1/10 the size.

댓글 수: 1

Andrew Reibold
Andrew Reibold 2014년 10월 15일
Its very easy to replace every single point with the average too if thats what you need. (Same size, but every set of 10 numbers are all the same averaged number.)

댓글을 달려면 로그인하십시오.

Andrei Bobrov
Andrei Bobrov 2014년 10월 16일

1 개 추천

I want have 'vote' too!
out = accumarray(ceil((1:numel(x))'/10),x(:),[],@mean);

댓글 수: 2

b=blkproc(rand(5760,1),[10,1],@mean)
I can't think of a solution shorter than my reshape one in terms of Code Golf?
john creighton
john creighton 2014년 10월 16일
haha. yours works too...so well deserved 'vote' :)

댓글을 달려면 로그인하십시오.

john creighton
john creighton 2014년 10월 16일

0 개 추천

Both answers worked perfectly. Thanks guys. sorry can only accept one answer lol

카테고리

도움말 센터File Exchange에서 Language Fundamentals에 대해 자세히 알아보기

질문:

2014년 10월 15일

댓글:

2014년 10월 16일

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by