Append rows at the end of matrix for loop

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Rachel Ramirez
Rachel Ramirez el 7 de En. de 2021
Comentada: Rik el 24 de En. de 2021
Hello,
I need some help with a for loop. I have matrix A (5x2) and matrix B (3x2). (See image) I want to have 3 different combinations (3 new matrices size 6x2) and for each combination add one row from matrix B at the end of matrix A. How can I approach this?
  1 comentario
Rik
Rik el 24 de En. de 2021
Just in case you try to edit away your question I made a capture.
Why do you think your question is not clear? Why not improve it instead of flagging it?

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Respuestas (2)

Jan
Jan el 7 de En. de 2021
Editada: Jan el 7 de En. de 2021
A = rand(5, 2);
B = rand(3, 2);
Result = cell(1, 3);
for k = 1:3
Result{k} = [A; B(k, :)];
end

Robert Rasche
Robert Rasche el 7 de En. de 2021
It depends on the format/order in which the new matrices shall be output. Do you want them as a collection (cell) or one after another?
In any case, it should be pretty simple, because you just have iterate through the height-dimension of B:
C = cell(size(B, 1), 1); % in case a collection is needed (*)
for i in 1:size(B, 1)
M = [A; B(i, :)];
C{i} = M; % (*)
end
or something.
  2 comentarios
Rachel Ramirez
Rachel Ramirez el 7 de En. de 2021
Thank you for your reply.umm haven't used cell before, will look into what it's for.
What I wanted to output it's something like the following. 3 matrices. each of the matrices has the 5x2 values from matrix A and at the end 1/3 rows from Matrix B.
Also another thing, I'm using an example of an specifc size but since my matrix would vary i would rather not specify a size in specifc. is that possible?
Robert Rasche
Robert Rasche el 8 de En. de 2021
This is what that code will do and it will also work with different sizes - as long as A and B have the same number of columns of course.
The crucial part in both Jan's and my answer is where it says
= [A; B(i,:)];
  • snip out a part of B, the i-th line (via ...(i, :)) and then
  • stich it to your A by vertical concatenation (via [... ; ...])
If you want there to always be 3 result matrices instead of as many as B has rows, then the loop bounds and snipping-part would have to be ajusted, but i don't think it is what you want.

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