Replace efficiently values in matrix without for-loops

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Matteo
Matteo 2012-3-2
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

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采纳的回答

Matteo
Matteo 2012-3-2
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

更多回答（2 个）

the cyclist
the cyclist 2012-3-2
Simple and elegant:
b(A)
Jonathan Sullivan
B = b(A);

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