Non-Overlapping Moving Sum

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.

 Accepted Answer

Try this —
Rain = [1,2,3,4,5,6,7,8,9,10];
rRain = reshape(Rain, 2, [])
rRain = 2×5
1 3 5 7 9 2 4 6 8 10
sumRain = sum(rRain)
sumRain = 1×5
3 7 11 15 19
These could be combined into one line, however I kept them separate to demonstrate how it works.
.

6 Comments

Actually my matrix has size 135*129*118. I wanted to accomplish this non-overlapping 2 sum so that my final matrix is 135*129*59. I tried using reshape, but may be i did some mistake the resultant matrix wasn't the one i wanted, could you provide some hint for 3-d matrix?
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.
.
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:
You can download sepblockfun, which generalizes the technique Star Strider has shown to higher dimensions.
A=rand(135,129,118);
B=sepblockfun(A,[1,1,2],'sum');
whos A B
Name Size Bytes Class Attributes A 135x129x118 16439760 double B 135x129x59 8219880 double
@Matt J — Thank you! I didn’t think to look in FEX.
This worked thanks @Matt J and everyone

Sign in to comment.

More Answers (1)

Image Analyst
Image Analyst on 27 Jun 2021
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.

Products

Release

R2019a

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!