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Can I partition a matrix ito several seperated parts ?

조회 수: 2 (최근 30일)
omar th
omar th 2022년 3월 31일
댓글: omar th 2022년 4월 1일
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
Voss 2022년 3월 31일
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
  댓글 수: 4
Walter Roberson
Walter Roberson 2022년 4월 1일
Ah, I don't think there was any way any of us could have guessed you wanted to select in that order.
omar th
omar th 2022년 4월 1일
actually its not exactly in that order for sure i changed it, but this approach worked in my assumption
thanks again

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추가 답변 (1개)

Walter Roberson
Walter Roberson 2022년 3월 31일
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
Name Size Bytes Class Attributes A 29x1 232 double B 7x1 56 double C 7x1 56 double D 7x1 56 double E 8x1 64 double parts 4x1 648 cell
  댓글 수: 3
Walter Roberson
Walter Roberson 2022년 3월 31일
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
Name Size Bytes Class Attributes A 35x1 280 double B 8x1 64 double C 8x1 64 double D 8x1 64 double E 11x1 88 double each 1x1 8 double nrow 1x1 8 double parts 4x1 696 cell
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.
omar th
omar th 2022년 4월 1일
thank you so much

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