Can I partition a matrix ito several seperated parts ?
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can I partition this matrix into 4 parts ?
assume this matrix 29X1, A = [2;2;-3;4;5;6;7;-8;9;-6;5;4;-2;1;3;-9;8;7;4;-5;6;3;2;-1;4;-7;-8;5;-6];
so, I want to group this matrix into 4 seperated groups such as B=7X1, C=7X1, D=7X1, and E=8X1 ?
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Voss
2022 年 3 月 31 日
A = [2;2;-3;4;5;6;7;-8;9;-6;5;4;-2;1;3;-9;8;7;4;-5;6;3;2;-1;4;-7;-8;5;-6];
% maybe this:
B = A(1:7);
C = A(8:14);
D = A(15:21);
E = A(22:29);
% or maybe this:
B = A(2:4:end);
C = A(3:4:end);
D = A(4:4:end);
E = A(1:4:end);
% etc., many other partitionings are possible
4 件のコメント
Walter Roberson
2022 年 4 月 1 日
Ah, I don't think there was any way any of us could have guessed you wanted to select in that order.
その他の回答 (1 件)
Walter Roberson
2022 年 3 月 31 日
A = [2;2;-3;4;5;6;7;-8;9;-6;5;4;-2;1;3;-9;8;7;4;-5;6;3;2;-1;4;-7;-8;5;-6];
parts = mat2cell(A, [7, 7, 7, 8], 1);
[B, C, D, E] = parts{:};
whos
3 件のコメント
Walter Roberson
2022 年 3 月 31 日
Your desired outcome is not defined. Do you always want 4 blocks, and the first three of them should be equal, and the last should absorb any extra capacity ?
A = randi([-9 9], 35, 1);
nrow = size(A,1);
each = floor(nrow/4);
parts = mat2cell(A, [each, each, each, nrow - 3*each], 1);
[B, C, D, E] = parts{:};
whos
That gives 8, 8, 8, 11.
But perhaps you would instead prefer 9, 9, 9, 8 -- in which the last matrix might be shorter, but the distribution is more even.
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