Non-Overlapping Moving Sum
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Consider any matrix, say Rain=(1:10)'. I want to compute a non-overlapping moving sum with two element.
Rain=1,2,3,4,5,6,7,8,9,10
Compute=1+2,3+4,5+6,7+8,9+10
With traditional movsum command, the moving sum is overlapping elements. I want to evaluate non-overlapping moving sum. Any in-built function? I have tried with loops and all those, it works but in-built function would be fast to compute.
采纳的回答
Try this —
Rain = [1,2,3,4,5,6,7,8,9,10];
rRain = reshape(Rain, 2, [])
sumRain = sum(rRain)
These could be combined into one line, however I kept them separate to demonstrate how it works.
.
6 个评论
Suman Dhamala
2021-6-27
Star Strider
2021-6-27
It always helps to have the actual problem in the beginning.
Every combination of permute, reshape and sum that I experimented with failed to provide the desired result.
.
Image Analyst
2021-6-27
Star's right - avoids wasting time if we know the full situation in advance. However he showed you a nice trick that will come in useful. It's also in the FAQ:
Star Strider
2021-6-27
@Matt J — Thank you! I didn’t think to look in FEX.
Suman Dhamala
2021-6-27
更多回答（1 个）
Image Analyst
2021-6-27
0 个投票
Since you have an image, you can do it with blockproc. However it only works with 2-D arrays so you'll have to do it once on each slice, then again along the z direction. I'm attaching some blockproc demos. I haven't done it with a 3-D image so you're on your own but I'm pretty sure it can be done.
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