Matlab: dividing a vector into overlapping pieces of a fixed size

Question

I have a vector that I would like to split into overlapping subvectors of size cs at shifts. Imagine an input vector:

 v=[1 2 3 4 5 6 7 8 9 10 11 12 13]; % A=[1:13] 

given a chunksize of 4 ( cs=4 ) and shift 2 ( sh=2 ), the result should look like this:

 [1 2 3 4] [3 4 5 6] [5 6 7 8] [7 8 9 10] [9 10 11 12] 

note that the input vector is not necessarily divided by chunksize , and therefore some subvections are discarded. Is there any quick way to figure this out without the need for use, for example. a for loop? In a related post, I found how to do this, but when looking at non-overlapping subvectors.

Answer 1

You can use the bsxfun function as follows:

 v=[1 2 3 4 5 6 7 8 9 10 11 12 13]; % A=[1:13] cs=4; sh=2; A = v(bsxfun(@plus,(1:cs),(0:sh:length(v)-cs)')); 

Here's how it works. bsxfun applies some basic functions on 2 arrays and performs some repmat like if the input data sizes are not suitable. In this case, I generate the indices of the first block and add the offset of each chunck. Since one input is a row vector and the other is a column vector, the result is a matrix. Finally, when indexing a vector with a matrix, the result is a matrix, which is exactly what you expect.

And this one-line (almost) always fun :).

Answer 2

I guess the easiest way is actually a loop. The decision to vectorize may be faster, but if the result is properly pre-distributed, the loop should also run decently.

 v = 1:13 cs = 4; sh = 2; myMat = NaN(floor((numel(v) - cs) / sh) + 1,cs); count = 0; for t = cs:sh:numel(v) count = count+1; myMat(count,:) = v(t-cs+1:t); end 

Answer 3

Do you have a signal processing toolbar? Then the buffer command. First look at the bare output:

 buffer(v, 4, 2) ans = 0 1 3 5 7 9 11 0 2 4 6 8 10 12 1 3 5 7 9 11 13 2 4 6 8 10 12 0 

This is clearly the right idea, with a little tweaking necessary to give you exactly what you want:

 [yz] = buffer(v, 4, 2, 'nodelay'); y.' ans = 1 2 3 4 3 4 5 6 5 6 7 8 7 8 9 10 9 10 11 12 

However, consider leaving vectors in half, as this is better suited to most use cases. For example, the average for each window is simply mean matrix, because the default column is.

Answer 4

You can accomplish this with ndgrid :

 >> v=1:13; cs=4; sh=2; >> [Y,X]=ndgrid(1:(cs-sh):(numel(v)-cs+1),0:cs-1) >> chunks = X+Y chunks = 1 2 3 4 3 4 5 6 5 6 7 8 7 8 9 10 9 10 11 12 

The good thing about the second syntax of the colon operator ( j:i:k ) is that you do not need to calculate k exactly (for example, 1:2:6 gives [1 3 5] ) if you plan to undo additional entries, as in this task. It automatically goes into j+m*i , where m = fix((kj)/i) ;

Various tests:

 >> v=1:14; cs=5; sh=2; % or v=1:15 or v=1:16 >> [Y,X]=ndgrid(1:(cs-sh):(numel(v)-cs+1),0:cs-1); chunks = X+Y chunks = 1 2 3 4 5 4 5 6 7 8 7 8 9 10 11 10 11 12 13 14 

And a new line will be formed using v=1:17 . Does this support all cases as needed?

Answer 5

How about this? First, I generate starting indices based on cs and sh for clipping individual vectors outside the full-sized vector, then I delete all indices for which idx+cs will exceed the length of the vector, and then I cut from the individual subvectors via arrayfun and then transforming them into a matrix :

 v=[1 2 3 4 5 6 7 8 9 10 11 12 13]; % A=[1:13] cs=4; sh=2; idx = 1:(cs-sh):length(v); idx = idx(idx+cs-1 <= length(v)) A = arrayfun(@(i) v(i:(i+cs-1)), idx, 'UniformOutput', false); cell2mat(A') 

eg. for cs=5; sh=3; cs=5; sh=3; this will give:

 idx = 1 3 5 7 ans = 1 2 3 4 5 3 4 5 6 7 5 6 7 8 9 7 8 9 10 11 

Depending on where the cs; sh values ​​come from cs; sh cs; sh , you probably want to introduce simple error checking so that cs > 0; as well as sh < cs . sh < 0 would be theoretically possible if you want to leave some values ​​between them.

EDIT : A very small bug has been fixed, now it should work for different combinations of sh and cs.