Fastest way to substitute elements in a matrix at given positions?

1 Ansicht (letzte 30 Tage)
Good evening,
suppose I have an array A with size (n,m) and an array B with size (n,l), with l<m.
Suppose elements of array B point to elements of array A in the following fashion:
  • B(x,y) --> A(x, B(x,y))
What is the fastest way to substitute all elements of A that are pointed by elements of B with a given value? At present I am using for loops but I guess they are not very efficient.
Also, would it be different if matrix A would be generated by repetition of the same known row n times?
Thanks a lot for any help or hint you are willing to give!!

Akzeptierte Antwort

Matt J
Matt J am 15 Okt. 2019
idx = sub2ind([n,m], repmat((1:n).',1,l) ,B);
A(idx)=value;
  5 Kommentare
Matt J
Matt J am 16 Okt. 2019
Bearbeitet: Matt J am 16 Okt. 2019
It depends on what the code for the loop looks like. Generally speaking though, Matlab's JIT compiler can do some things to optimize M-coded for-loops, but is limited by the complexity of operations done inside the loop and your Matlab version, of course.
If you can avoid M-Coded for-loops, as my solution does, it often leads to better performance because it doesn't rely on the JIT. It is just running pre-compiled C/C++ code internally.

Melden Sie sich an, um zu kommentieren.

Weitere Antworten (0)

Kategorien

Mehr zu Matrix Indexing finden Sie in Help Center und File Exchange

Produkte


Version

R2019a

Community Treasure Hunt

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

Start Hunting!

Translated by