How to call index of vector in matrix?
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ha ha
el 27 de Ag. de 2017
Comentada: ha ha
el 27 de Oct. de 2017
Hello, I have: n-by-3 matrix A, and n-by-1 matrix B:
A=[ x1 y1 z1
x2 y2 z2
x3 y3 z3
x4 y4 z4
.......
xn yn zn ]
B=[ 3
2
7
1
...
n ]
B is index (labelling) matrix of A.
I wanna assign the vector A to vector B.
Ex:
(x1 y1 z1) assign to 1
(x3 y3 z3) assign to 3
......................
(xn yn zn) assign to n
- Instead of working with matrix A, I can work with "labelling" matrix B.
and then,
+ when I call 3 in matrix B, it will show the value (x3 y3 z3) from matrix A.
+ when I call 7 in matrix B, it will show the value (x7 y7 z7) from matrix A.
+ when I call [3,7] in matrix B, it will show the value matrix:
A=[ x3 y3 z3
x7 y7 z7]
+ when I call n in matrix B, it will show the value (xn yn zn) from matrix A. ...............................................................
and vice versa ( I call (x3 y3 z3) from A, it will show 3 in B.....)
- How to write a code to call matrix A from matrix B , and vice versa?
(the number 3,2,7,1,....n: in matrix B is are arbitrary numbers and x1,x2,x3...xn : are the number)
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Respuesta aceptada
Image Analyst
el 27 de Ag. de 2017
I don't know what it means to "call label 3" or B. Is B always 1,2,3,4 etc. - the natural counting numbers? Or can they be arbitrary numbers? If you have a number 3, and want to look up B(3) and use that as an index to get that row of A, you can simply do
out = A(B(3), :);
and if you want to "show" it, you can just leave off the semicolon:
out = A(B(3), :)
or use fprintf() :
fprintf('A(%d) = [%f, %f, %f]\n', B(3), A(B(3), 1), A(B(3), 2), A(B(3), 3));
or msgbox() or helpdlg():
message = sprintf('A(%d) = [%f, %f, %f]\n', B(3), A(B(3), 1), A(B(3), 2), A(B(3), 3));
uiwait(helpdlg(message));
Is that what you want?
6 comentarios
Jan
el 27 de Oct. de 2017
@ha ha: Please use flags only to inform admins and editors about messages, which conflict with the terms of use of this forum. Thanks.
Más respuestas (1)
Andrei Bobrov
el 27 de Ag. de 2017
Editada: Andrei Bobrov
el 27 de Ag. de 2017
A = [...
6 -3 4
7 8 -5
-4 8 6
7 1 8
3 6 4
-4 -4 5
-2 0 5
2 7 0
8 6 4
8 8 -3];
A2B(A,[6 -3 4;3 6 4])
A2B(A,[1;2])
here function A2B:
function out = A2B(A,in)
if size(in,2) == 1
out = A(in,:);
else
out = find(ismember(A,in,'rows'));
end
end
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