creating matrix using output elements

Hi, i havecode below which ouputs given values independently. however, i want the ouputs to be under one matrix,what can i add to the code to do this job. B is an n by 1 matrix say B=[1:1:24]'. How can I collect all the values of Hourly_mean to form one matrix? kindly assist. Thank you

B=[1:1:576]'

for k=1:1:numel(B)

Hourly_mean=mean(B(k:24:end))

end

[Hourly_mean]

Answer 1

[Fangjun Jiang]

B=[1:1:576]';
mean(reshape(B,24,[]))

ans = 1×24

12.5000 36.5000 60.5000 84.5000 108.5000 132.5000 156.5000 180.5000 204.5000 228.5000 252.5000 276.5000 300.5000 324.5000 348.5000 372.5000 396.5000 420.5000 444.5000 468.5000 492.5000 516.5000 540.5000 564.5000

[okoth ochola]

Hi, thank you for your answer however reshape() function splits the matrix by picking the first 24 values consecutively, while my questions requires the code to pick the first value then the second value in steps of 24 then it picks another till the end of the matrix. Do you have any ideas?

[Steven Lord]

Let's take a smaller example that demonstrates the technique. Say I want to take the mean of every 6th element of B. We can reshape B into a matrix.

B = 1:24;
C = reshape(B, 6, 4)

C = 6×4

1 7 13 19 2 8 14 20 3 9 15 21 4 10 16 22 5 11 17 23 6 12 18 24

Now take the mean along the 2nd dimension.

D = mean(C, 2)

D = 6×1

10 11 12 13 14 15

Spot check that D is correct by manually computing the mean of the 3rd, 9th, 15th, and 21st element of B. Does that match D(3)?

sum(B(3:6:24))./4

ans = 12

[Fangjun Jiang]

@Steven Lord, good catch! mean(C,2) is more likely the needed outcome than mean(C).

Answer 2

[Image Analyst]

Try splitapply, grpstats, or groupsummary