Vectorized "find an replace"

I have a matrix `A` and an array `map`, both consist of integers. I would like to replace every entry where `A == i` with `map(i)`. I wrote the following non-vectorized code

Aorig = A;
for i = 1:numel(map)
  A(Aorig == i) = map(i);
end

Is there a good way to vectorize it?

Answer 1

[Guillaume]

Assuming all values in A are valid indices for map, simply:

A = map(A);

Answer 2

[Image Analyst]

Well clearly A=map(Aorig) won't work. You have to use intlut(). Here's an example:

% Create sample data
Aorig = int16(randi(9, 20, 1))-1
% Initialize look up table for 16 bit signed integers:
map = zeros(65536, 1, class(Aorig));
% Now, enter your custom values at the proper locations.
% Map value for 0 should happen at index 32769 of map.
% Adjust positions to that that's true.
map((0:8)+32769) = int16([-9999,-8,-7,-6,-5,-4,-3,-2,-1])  % Just an example!
% B = Aorig(map) % Throws error.
A = intlut(Aorig, map)

This works and works not only for positive integers but for negative integers as well. Adjust map elements to create the lookup table that you want. I just used an arbitrary example.

[Guillaume]

"Well clearly A=map(Aorig) won't work"

Why not? In the original code, i goes from 1 to numel(map). The only case where map(A) wouldn't replace that is when numel(map) < max(A(:)). But in that case, some original values in A would not get replaced, so it's dubious that it'd be the case.

[Image Analyst]

It doesn't work for all values of A and map, for example negative or zero values. It doesn't work for values for where map is greater than the A either. For an all positive example where it doesn't work:

% Create 10 sample elements in the range [1, 3]
Aorig = int16(randi([1, 3], 10, 1))
% As an example, replace these A values with these replacement values.
% Replace 1's with 99, 2's with 0, 3's with 73.
map = int16([99, 0, 73])  % Just an example!
B = Aorig(map) % Throws error.

I got the impression the range of the A could be any integers, and the desired replacement values, in map, could also be any values that he desired to replace them with. My code will work for those conditions, I believe, while B=Aorig(map) doesn't.

[Guillaume]

Well, clearly, from the code provided, the look-up values are always strictly positive. As I said, it wouldn't work for look-up values greater than the number of elements in map but it wouldn't make much sense to replace only part of the look-ups.

Note that the values of map are irrelevant. They can be positive, 0, negative, integer or non-integer.

In the OP case, I do believe that

map(A)  %and not the opposite

would be the simplest solution.