Extract consecutive elements from a vector in a sliding manner
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Hi all,
If I have a vector:
A = [1 ;2 ;3 ;4 ;5 ;6];
and I want to extract 2 elements in a way that I should get in each loop:
for i 1:size(A,1)
A_slice = [?];
end
The answer I'm looking for, for each loop I should get:
%Loop 1: A_slice=[1;2]
%Loop 2: A_slice=[3;4]
%Loop 3: A_slice=[5;6]
any help would be appreicted.
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採用された回答
Matt J
2025 年 4 月 30 日
A = [1 ;2 ;3 ;4 ;5 ;6];
Ar=reshape(A,2,[]);
for i=1:size(Ar,2)
A_slice=Ar(:,i)
end
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その他の回答 (2 件)
dpb
2025 年 4 月 30 日
編集済み: dpb
2025 年 4 月 30 日
Alternatively, depending upon end use, without the explicit loop can generate the subsets directly...
x=[1:6].';
n=2;
slice=arrayfun(@(i)x(i:i+n-1),1:n:numel(x)-1,'UniformOutput',false)
or, more succinctly
slice=num2cell(reshape(x,n,[]),1)
ADDENDUM: I see now that the latter above was already shown by @Matt J, I just recast into the cell array with generic number of elements. Of course, either fails if numel(x) isn't integer multiple of n.
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Walter Roberson
2025 年 4 月 30 日
A = [1 ;2 ;3 ;4 ;5 ;6];
A_slice = buffer(A, 2)
now do things with the columns.
buffer() requires the Signal Processing Toolbox. (Personally, I think it ought to be made part of basic MATLAB.)
buffer() has the advantage of smoothly handling the case where the input vector is not an exact multiple of the buffer size.
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