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Replacing n-d array elements based on pre-defined list

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Hi,
Suppose I have an n-d array of q unique integers. I want to replace the value of every element in this n-d array based on a list (q by 2 matrix) like this:
1 k
2 l
3 m
...,
which means that every element with value 1 in the n-d array is replaced by k, every element with value 2 is replaced by l, etc. k, l, m... are double-precision numbers.
What's the most efficient way of doing this? Obviously, one could loop through 1, 2, 3..., find their linear indices and set elements with those indices to k, l, m... I would like to do this faster, if possible.
Any ideas are appreciated.

Accepted Answer

Matt J
Matt J on 11 Jun 2013
Edited: Matt J on 11 Jun 2013
v=[k l m ...];
X=...%Your n-D array of integers
result = v(X);

More Answers (1)

Roger Stafford
Roger Stafford on 11 Jun 2013
Edited: Roger Stafford on 11 Jun 2013
If it isn't assumed that vector 'l' contains successive integers starting with 1, then you could do this:
[tf,loc] = ismember(V(:),l);
V(tf) = k(loc(tf));
where 'k' is the vector of values to replace those in 'l' and where 'V' is the n dimensional array. ('l' is the lowercase 'L', not the numeral.)
  2 Comments
Pal
Pal on 11 Jun 2013
Thanks. You mean l(:,1), the integer elements, I presume.
Roger Stafford
Roger Stafford on 11 Jun 2013
My apologies. I misread your two vectors as being named 'l' and 'k'. My code still works whatever you name them provided they are the same length. It doesn't matter whether they contain integers or non-integers. Whenever an element in V is equal to one in 'l' (or whatever you call it,) it is replaced by the corresponding element in 'k' (or whatever you call that one.)

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