# Breaking and spliting vector into groups and averaging

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Albert am 10 Mär. 2021
Bearbeitet: Albert am 11 Mär. 2021
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];
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### Akzeptierte Antwort

Adam Danz am 10 Mär. 2021
Bearbeitet: Adam Danz am 10 Mär. 2021
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
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Adam Danz am 11 Mär. 2021
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 am 11 Mär. 2021
Bearbeitet: Albert am 11 Mär. 2021
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)

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