Array manipulation: Suppose you have an array.How to take windows with overlapping along column and repeat to next doing same

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MSP 2017년 1월 12일
댓글: Kirby Fears 2017년 1월 23일
for example a= [1 2 3 4 5; 4 7 9 1 5; 8 7 3 2 1; 3 2 9 1 7]
so what is required to be done is take out a window of size 3 and shift the window 1 data point in the next iteration. SO the result will show
1 4 8
then 483 for the first column in 1st column of a new array.So the first array would have 6 elements and it will be a 6*5 array.
so the new array would have first two columns like 1st column [1;4;8;4;8;3] 2nd column [2;7;7;7;7;2]

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Kirby Fears 2017년 1월 12일
편집: Kirby Fears 2017년 1월 12일
Indexing and stacking as shown below will work for your example. Does this work for your real use case also?
a = [1 2 3 4 5; 4 7 9 1 5; 8 7 3 2 1; 3 2 9 1 7];
b = [a(1:end-1,:); a(2:end,:)];
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MSP 2017년 1월 12일
편집: MSP 2017년 1월 12일
Thanks for the help, works for the example but would have rather wanted a general solution like when the window size gets bigger as well as the shift ;consider window 128,shift 64.The main objective is to get windows in same position of two different arrays like 'a' and then compare to find out the relationships.maybe looping can work too.if u got any better ideas pls share.
Kirby Fears 2017년 1월 23일
The example I posted is easily extended to general window and shift sizes. For example, my original solution can be written with window size 3 and step size 1.
windowSize = 3;
stepSize = 1;
a = [1 2 3 4 5; 4 7 9 1 5; 8 7 3 2 1; 3 2 9 1 7];
b = [a(1:windowSize,:); a(1+stepSize:windowSize+stepSize,:)];

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