Average of evey nth row of a large matrix

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This is my problem:
I have a matrix, let's say A(5000,10). So, row 1 of matrix A is comparable to row 101, row 2 is comparable to row 102 and so on. Each 100 of those sets are a bin of data (radial distribution function to be specific).
Now, I need to match up these rows and calculate their average value. So I will have a matrix of 50x10. Those 50 is the average of all the nth element (1, 101, 201 and so on). Basically I will need the average of all the bins.

Accepted Answer

Dave B
Dave B on 11 Aug 2021
To average every nth row:
a=rand(100,10);
n = 10;
mean(a(1:n:height(a),:),2)
ans = 10×1
0.7132 0.5279 0.5908 0.5117 0.4052 0.5002 0.4536 0.5225 0.4116 0.4415
But I think you want the average of the first block of n rows, the second block of n rows, etc. An easy way to do this is by making a little index of which rows should go into the average and then using groupsummary. This has a bonus that it's really extensible - when you later want the std of each block, it's trivially easy:
ind=floor(((1:height(a)) - 1)/n)+1 % an index of n ones, n twos, etc.
ind = 1×100
1 1 1 1 1 1 1 1 1 1 2 2 2 2 2 2 2 2 2 2 3 3 3 3 3 3 3 3 3 3
b=groupsummary(a,ind','mean')
b = 10×10
0.4956 0.5382 0.4825 0.5698 0.4879 0.4804 0.4992 0.7141 0.6146 0.3100 0.4091 0.6328 0.6922 0.4547 0.4437 0.4993 0.4103 0.6151 0.4009 0.5329 0.5224 0.6052 0.5603 0.2870 0.5020 0.4421 0.6260 0.5311 0.5578 0.5436 0.4771 0.3879 0.3807 0.4743 0.2958 0.3670 0.5435 0.5268 0.5284 0.4745 0.4394 0.4199 0.6134 0.3321 0.4016 0.5233 0.4660 0.5426 0.5008 0.5224 0.3314 0.7157 0.4940 0.7073 0.5422 0.4870 0.4935 0.5722 0.6656 0.4735 0.4324 0.6773 0.3981 0.5952 0.4559 0.3111 0.6122 0.3747 0.6102 0.5560 0.5078 0.5559 0.5401 0.3599 0.5674 0.4305 0.5721 0.5610 0.4347 0.4187 0.5629 0.4248 0.6734 0.3919 0.4130 0.6040 0.3570 0.6186 0.4929 0.5233 0.3555 0.3813 0.4285 0.5278 0.4954 0.5590 0.5031 0.4028 0.3891 0.3415
% check that it's correct:
isequal(mean(a(1:10,:)),b(1,:))
ans = logical
1
isequal(mean(a(11:20,:)),b(2,:))
ans = logical
1
  4 Comments
Dave B
Dave B on 12 Aug 2021
I'm not sure if I read that correctly, but I would expect the last line to be:
X = sum(b,3);
You want to some across the third dimension.

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