How to average first two dimension of 3D array/matrix?

20 次查看(过去 30 天)
Hi all,
I am calculating the average of first two dimension of a 3D matrix but it is showing following error:
Error using sum
Dimension argument must be a positive integer scalar within indexing range.
Error in mean (line 116)
y = sum(x, dim, flag) ./ size(x,dim);
Also, I have used a simplest example given in topic 'Mean of Array Page' in https://in.mathworks.com/help/matlab/ref/mean.html but I am getting same error.
I am using MATLAB 2018a
Please help!
Thanks in advance
  2 个评论
UTKARSH VERMA
UTKARSH VERMA 2021-9-1
I have used a simplest example given in topic 'Mean of Array Page' in https://in.mathworks.com/help/matlab/ref/mean.html and I got an error:-
""Error using sum
Dimension argument must be a positive integer scalar within indexing range.
Error in mean (line 116)
y = sum(x, dim, flag) ./ size(x,dim);""
I guess my Matlab version is older (i.e. 2018a)

请先登录,再进行评论。

采纳的回答

Chunru
Chunru 2021-9-1
编辑:Chunru 2021-9-1
a = randi(3, 3, 4, 2)
a =
a(:,:,1) = 3 3 1 3 3 1 2 3 3 1 3 1 a(:,:,2) = 3 1 1 2 1 1 2 2 1 3 1 1
c = mean(reshape(a, [], size(a,3)))
c = 1×2
2.2500 1.5833

更多回答(2 个)

Wan Ji
Wan Ji 2021-9-1
编辑:Wan Ji 2021-9-1
Just use mean function to average first two dimension of 3D matrix
a = rand(2,3,4); % I just use rand function so the result is randomly displayed
b = mean(a,1:2) % 1:2 means the first two dimension
The result is
b = mean(a,1:2)
b(:,:,1) =
0.5258
b(:,:,2) =
0.2759
b(:,:,3) =
0.6214
b(:,:,4) =
0.4449
If you want to get array of b
Then
b = squeeze(b) % or b = b(:);
So
b =
0.5258
0.2759
0.6214
0.4449
  1 个评论
UTKARSH VERMA
UTKARSH VERMA 2021-9-1
Thank you for replying.
I guess my Matlab version is older (i.e. 2018a) due to which I am not able to get my result and it's showing same problem.

请先登录,再进行评论。


Steven Lord
Steven Lord 2021-9-1
The ability to specify a vector of dimensions over which to sum an array was introduced in release R2018b as stated in the Release Notes.

类别

Help CenterFile Exchange 中查找有关 Logical 的更多信息

产品


版本

R2018a

Community Treasure Hunt

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

Start Hunting!

Translated by