Divide a Large Matrix into Smaller Matrices in a Loop

2017 11 21
2 个回答
回答已采纳
更新时间：2017 11 21
6 次查看（30 天）

0 个投票

Hi all,
I have a Matrix of dimensions 20x5.
I need to run calculations, using For loops, on a program that first needs the first 4 lines of the large matrix, then the next 4 lines etc...
In other terms: 1 iteration ---> Large matrix lines [1 to 4 x 5] .... used for the calculations. Then 2 iteration ---> Large matrix lines [5 to 8 x 5] .... used for 2nd iteration calculations. Then 3 iteration ---> Large matrix lines [9 to 12 x 5].... used for 3rd iteration calculations etc...
Does anyone has an idea on how to make it happen?

请先登录，再回答此问题。

 采纳的回答

Stephen23
Stephen23 2017-11-21
编辑：Stephen23 2017-11-21

1 个投票

You can define the indexing quite easily:
mat = rand(20,5);
row = 4; % rows per group
grp = size(mat,1)/row;
for k = 1:grp
idr = (1:row)+(k-1)*row;
mat(idr,:)
end
Or split the data into a cell array using mat2cell:
C = mat2cell(mat,row*ones(1,grp),size(mat,2));
for k = 1:grp
C{k}
end

4 个评论
显示 2更早的评论 隐藏 2更早的评论

Nicolas Jalbert
Nicolas Jalbert 2017-11-21
编辑：Nicolas Jalbert 2017-11-21
Stephen,
Thanks for your answer. I need to use the first one, since I need the matrices for the calculations.
One question remains: would it be possible to "store" this 5 news matrices created? ---> In the program I would then call one small matrix at a time via let's say: mat(1), then I call mat(2) ... This would render the program more simple since I could perform this matrix division before entering the calculations.
I would then only include in the calculations a for loop calling each of the small matrix at a time.
Would it be possible?
Stephen23
Stephen23 2017-11-21
编辑：Stephen23 2017-11-21
"Would it be possible?"
Yes. That is exactly what the second method shows you how to do: C is a cell array containing all of the submatrices. Access them using cell array indexing:
C{1}
C{2}
etc
Or, if you really want to use indexing rather than mat2cell, then just put them into a cell array explicitly:
C = cell(1,grp);
for k = 1:grp
idr = (1:row)+(k-1)*row;
C{k} = mat(idr,:);
end
Nicolas Jalbert
Nicolas Jalbert 2017-11-21
I found a way without having to use cells arrays, this way we keep the matrices up for use. Because if they are stored in a cell array, we need a conversion (cell2mat?) to perform calculations with them right?
Here is my program:
mat = rand(20,5)
row = 4 % rows per group
grp = size(mat,1)/row
for k = 1:grp
idr = (1:row)+(k-1)*row;
mat(idr,:);
Matrix(:,:,k) = mat(idr,:);
end
Matrix
Is this a proper way to store these matrices? Or is there more elegant?
Stephen23
Stephen23 2017-11-21
编辑：Stephen23 2017-11-21
"Because if they are stored in a cell array, we need a conversion (cell2mat?) to perform calculations with them right?"
Not at all. You just need to use indexing, or something like cellfun.
"Is this a proper way to store these matrices?"
What is your definition of "proper"? The best way to store data is determined by the nature of that data and how is going to be processed, neither of what you have said anything about. Keeping that data in one numeric array will probably make numeric calculation a lot simpler, as you could write vectorized code.
"Or is there more elegant?"
An elegant way to rearrange that matrix into a 3D array is to use reshape and permute, no loop is required:
out = permute(reshape(mat.',[],row,grp),[2,1,3])

请先登录，再进行评论。

更多回答（1 个）

Andrei Bobrov
Andrei Bobrov 2017-11-21
编辑：Andrei Bobrov 2017-11-21

1 个投票

mat1 - matrix with size (20,5)
row1 = 4;
Matrix1 = permute(reshape(mat1.',size(mat1,2),row1,[]),[2,1,3]);

类别

在 帮助中心 和 File Exchange 中查找有关 Matrices and Arrays 的更多信息

标签

Community Treasure Hunt

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

Start Hunting!

Translated by