creating matrix using output elements

4 次查看（过去 30 天）
okoth ochola
okoth ochola 2023-1-19
评论： Fangjun Jiang 2023-1-20
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]

请先登录，再回答此问题。

采纳的回答

Fangjun Jiang
Fangjun Jiang 2023-1-19
Ran in:
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
  3 个评论
显示 1更早的评论
Steven Lord
Steven Lord 2023-1-20
Ran in:
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
Fangjun Jiang 2023-1-20
@Steven Lord, good catch! mean(C,2) is more likely the needed outcome than mean(C).

请先登录，再进行评论。

更多回答（1 个）

Image Analyst
Image Analyst 2023-1-20

类别

在 Help Center 和 File Exchange 中查找有关 MATLAB 的更多信息

标签

产品


版本

R2016a

Community Treasure Hunt

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

Start Hunting!

Translated by