Replace efficiently values in matrix without for-loops

2 views (last 30 days)
Hi, This is my problem. Suppose you have a vector b:
b=[234 786 674 23];% the values are chosen at random for this example
and a matrix A:
A=[2 1;
4 4]; %the value of the elements of A should be <=length(b)
it exists a method or a function to obtain a second matrix B of the same dimensions of A but with the values replaced by the corresponding value in b?? For the previous example:
B=[786 234;
23 23]; % i.e., B(i,j)=b(A(i,j))
this is easily to do with at least one for-loop. but I am looking for a function or method that does this without for loops, maybe in a vectorized way (thus in an efficient way)!
Can someone help me??
Thank you,
Matteo

Accepted Answer

Matteo
Matteo on 2 Mar 2012
ok sorry, it was a stupid question.
I hadn't noticed that b(A) does exactly what I wanted :)
Hope anyway this could be useful for other people.
Bye, Matteo

More Answers (2)

the cyclist
the cyclist on 2 Mar 2012
Simple and elegant:
b(A)

Jonathan Sullivan
Jonathan Sullivan on 2 Mar 2012

Categories

Find more on Loops and Conditional Statements in Help Center and File Exchange

Community Treasure Hunt

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

Start Hunting!