Splitting a matrix based on certain values in the rows

17 次查看（过去 30 天）
Mr. 206
Mr. 206 2019-2-19
评论： Jos (10584) 2019-2-20
I have a matrix A like this:
A = [911 911;
0 2;
8 5;
7 3;
911 911;
5 3;
1 6;
6 7;
911 911;
3 5;
8 4];
I want to split the matrix A into three matrices (A1,A2,A3) based on the row values 911 like this:
A1 = [0 2; 8 5; 7 3];
A2 = [5 3; 1 6; 6 7];
A3 = [3 5; 8 4];
I need to do this thing inside a for loop which will give the spitted matrix one after another.
  3 个评论
显示 1更早的评论
Mr. 206
Mr. 206 2019-2-19
This is just a dummy. It can be A, B,C or any other name.
Jan
Jan 2019-2-19
@Atta: A,B,C, ... suffers from exactly the same problems as A1, A2, A3, ... With using a cell array and and index, the code is fast and flexibel. You can e.g. simply run it for 12'781'986 rows without getting mad while typing the code with manually hidden names of variables.

请先登录，再进行评论。

请先登录，再回答此问题。

采纳的回答

Stephen23
Stephen23 2019-2-19
编辑：Stephen23 2019-2-19
No loop required:
>> idx = cumsum(all(A==911,2));
>> row = 1:numel(idx);
>> fun = @(r){A(r(2:end),:)};
>> C = accumarray(idx,row(:),[],fun);
>> C{:}
ans =
0 2
8 5
7 3
ans =
5 3
1 6
6 7
ans =
3 5
8 4
  4 个评论
显示 2更早的评论
Mr. 206
Mr. 206 2019-2-20
编辑：Mr. 206 2019-2-20
Thanks, I got it.
I just need another help, I have another 5 by 6 matrix (say D). I want to pad the matrices C{1}, C{2} and C{3} by some random number, so that the row numbers are equal.
For example, C{3} has 2 rows, so i need to put three e.g 911911 value at the bottom to make the row number equal to matrix D (5 by 6)
Jos (10584)
Jos (10584) 2019-2-20
% Pad each cell with rows of 911,
% so that it has the same number of rows as D
% using CELLFUN with an anonymous padding function using REPMAT
C2 = cellfun(@(m) [m ; repmat(911, size(D,1)-size(m,1),size(m,2))], C)

请先登录，再进行评论。

更多回答（1 个）

Jan
Jan 2019-2-19
编辑：Jan 2019-2-19
A = [911 911;
0 2;
8 5;
7 3;
911 911;
5 3;
1 6;
6 7;
911 911;
3 5;
8 4];
index = [find(A(:, 1) == 911); size(A, 1) + 1];
n = numel(index) - 1;
Result = cell(1, n);
for k = 1:n
Result{k} = A(index(k):index(k+1)-1, :); % [EDITED]
end
  2 个评论
Mr. 206
Mr. 206 2019-2-19
It's throwing an error, like "error: A(I,J): row index out of bounds; value 12 out of bound 11"
Jan
Jan 2019-2-19
@Atta: Yes, I had a typo in my code. It is fixed now. Sometimes I expect the readers to fix bugs, when they are not too hard.
The result is a cell array and you access it as Result{1}. This is much smarter than hiding an index in the name of a variable.

请先登录，再进行评论。

类别

在 Help Center 和 File Exchange 中查找有关 Logical 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