Splitting a matrix based on certain values in the rows
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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 commentaires
Jan
le 19 Fév 2019
@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.
Réponse acceptée
Stephen23
le 19 Fév 2019
Modifié(e) : Stephen23
le 19 Fév 2019
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 commentaires
Jos (10584)
le 20 Fév 2019
% 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)
Plus de réponses (1)
Jan
le 19 Fév 2019
Modifié(e) : Jan
le 19 Fév 2019
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 commentaires
Jan
le 19 Fév 2019
@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.
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