Logical Indexing via multiplication

20 views (last 30 days)
Hi,
I have a 3 x 3 matrix, B = [ 1 2 3; 4 5 6; 7 8 9 ]
I have a matrix, A = [ 0 1 0 ]'.
How can I extract only the middle row of matrix B, ie. [ 4 5 6] ?
If I multiply, I get the first and third row with zeros.
This is a simplification of a larger problem. How can I do it via multiplication of A and B?
I want all the rows of B that correspond where the index A equals 1.
Any help would be appreciated!
Thank you,
Inna
  1 Comment
Stephen23
Stephen23 on 21 Dec 2020
Basic logical indexing:
B = [1,2,3;4,5,6;7,8,9];
A = [0;1;0];
C = B(A==1,:)
C = 1×3
4 5 6

Sign in to comment.

Accepted Answer

David Goodmanson
David Goodmanson on 21 Dec 2020
Edited: David Goodmanson on 21 Dec 2020
Hi Inna, not done by multiplication, but:
ind = find(A==1)
rows_you_want = B(ind,:)
the colon means to take every column in whatever rows are selected.
  3 Comments
Inna Pelloso
Inna Pelloso on 21 Dec 2020
Thank you. For some reason, I ALWAYS forget about using the find function.

Sign in to comment.

More Answers (1)

Walter Roberson
Walter Roberson on 21 Dec 2020
Edited: Walter Roberson on 21 Dec 2020
This is a task that cannot be done by multiplication.
If you use .* elementwise multiplication then the size of the result is max() of the sizes of the inputs provided they are compatible sizes. Saying max() takes into account implicit expansion. The size of output never depends on the content of the data when you use .*
If you use A*B then size(A, 2) must equal size(B, 1) and the size of the output is always size(A, 1) by size(B, 2) no matter what the content of the variables are.
You can create projection matrices that select specific rows, but the size of the matrices depend on the number of rows being selected, so they have to be constructed outside of plain matrix multiplication.
  1 Comment
Walter Roberson
Walter Roberson on 21 Dec 2020
P = eye(size(B))
P = P(logical(A), :)
C = P * B
But you cannot construct P using just matrix multiplication without making decisions based on the content of A

Sign in to comment.

Tags

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!