Breaking and spliting vector into groups and averaging

조회 수: 2 (최근 30일)
Albert
Albert 2021년 3월 10일
편집: Albert 2021년 3월 11일
Hi,
I have a long one-dimensional time vector that I want to split in chunks and average each chunk. I have another logical vector (same length) of 0s and 1s that changes the sign to signal each chunk transition. How could I easily use the information on the logical vector to select each chunk of the time vector?
Thanks!
timevec = [0 2 3 4 5 0 9 3 4];
chunk = [0 0 0 1 1 1 1 0 0];

채택된 답변

Adam Danz
Adam Danz 2021년 3월 10일
편집: Adam Danz 2021년 3월 10일
timevec = [0 2 3 4 5 0 9 3 4];
chunk = [0 0 0 1 1 1 1 0 0];
g = cumsum([1, diff(chunk)]~=0); % assuming 'chunk' is a row vector
groupMeans = splitapply(@mean, timevec, g)
groupMeans = 1×3
1.6667 4.5000 3.5000
  댓글 수: 5
Adam Danz
Adam Danz 2021년 3월 11일
To define groups by the chunk vector resetting to 0,
chunk = [0 0 0 1 1 1 1 0 0 0 0 1]
chunk = 1×12
0 0 0 1 1 1 1 0 0 0 0 1
g = cumsum(diff([1,chunk])==-1)
g = 1×12
1 1 1 1 1 1 1 2 2 2 2 2
For the second question, do you mean this?
timevec = [0 2 3 4 5 0 9 3 4];
chunk = [0 0 0 1 1 1 1 0 0];
cond = logical([1 1 1 1 0 0 1 1 1]);
X = timevec(cond);
Y = chunk(cond);
g = cumsum([1, diff(Y)]~=0);
groupMeans = splitapply(@mean, X, g)
Albert
Albert 2021년 3월 11일
편집: Albert 2021년 3월 11일
For the first question, great, I didn't think it was so easy the adaptation!
For the second question it's a bit trickier because the Y array, after applying the condition in cond may totally mess around the chunks computation in the lines later on (transitions may be lost after applying the condition!). As a workaround I computed the chunks for all the values and applied the condition afterwards when computing the mean through a custom function as:
timevec = [0 2 3 4 5 0 9 3 4];
chunk = [0 0 0 1 1 1 1 0 0];
cond = logical([1 1 1 1 0 0 1 1 1]);
g = cumsum([1, diff(chunk)]~=0);
groupMeans = splitapply(@(vec,con)mean(vec(con)), timevec, cond, g)

댓글을 달려면 로그인하십시오.

추가 답변 (0개)

카테고리

Help CenterFile Exchange에서 Histograms에 대해 자세히 알아보기

Community Treasure Hunt

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

Start Hunting!

Translated by