Mean value of a subarray

조회 수: 8(최근 30일)
EldaEbrithil 22 Aug 2020
댓글: EldaEbrithil 22 Aug 2020
Hi all
i have an array V=1x115 i want to create another array which contains the mean value neatly of the first 8 V values than of the subsequent 8 and so on... in pratical terms: mean(V(1:8)), mean(V(8:16)),...mean(V(112:end))
Thank you for the help
Regards
댓글 수: 2표시숨기기 이전 댓글 수: 1
EldaEbrithil 22 Aug 2020
I solved the problem thanks to your upper link, now i understand the logic for the moving average
Regards

댓글을 달려면 로그인하십시오.

채택된 답변

편집: Adam Danz 22 Aug 2020
The answer in this link computes segmented averages as you describe except instead of averaging 1:8, 8:15, ... , it averages 1:8, 9:16, ..., .
Solution
To replicate the edges, you just have to add three line.
The full solutions is shown in this demo.
data = 1:22; % Demo data
winSz = 5; % Window size; winSz=5 results in indices of (1:5, 5:9, 9:13, ...)
replications = repmat([2;ones(winSz-2,1)],ceil(numel(data)/(winSz-1)),1);
dataPrep = repelem(data(:),replications(1:numel(data)));
dataPrep(1) = []; % do not duplicate 1st datapoint
% Reshape dataPrep into matrix.
% NOTE: 'dataPrep' must contain a number of element divisibly by 'winSz'.
% Otherwise, 'dataPrep' will be padded with NaN values so that it is
% divisible by 'winSz'.
if rem(numel(dataPrep),winSz)>0
nanAppend = nan(winSz - rem(numel(dataPrep),winSz),1);
else
nanAppend = [];
end
dataMat = reshape([dataPrep; nanAppend], winSz, []);
movingAverage = mean(dataMat,1,'Omitnan');
Results
The original data:
data =
Columns 1 through 20
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
Columns 21 through 22
21 22
The blocked version where means are computed over each column
dataMat =
1 5 9 13 17 21
2 6 10 14 18 22
3 7 11 15 19 NaN
4 8 12 16 20 NaN
5 9 13 17 21 NaN
The means (within a 5-element window)
movingAverage =
3 7 11 15 19 21.5
댓글 수: 3표시숨기기 이전 댓글 수: 2
EldaEbrithil 22 Aug 2020
Ahahah xD

댓글을 달려면 로그인하십시오.

R2019a

Community Treasure Hunt

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

Start Hunting!

Translated by