Question

This is a fairly simple task I want to perform, but I can’t seem to figure out a way to do it. I’ve tried sortrows, reshaping, and other solutions, but none of them do exactly what I want.

Essentially, I have two vectors from the same range of values, of unequal lengths. Some of the values are equal, some are not. E.g.

A = [1 5 20 30 53 70 92]
B = [2 3 4 16 20 30 60 95 100]

What I want to do is add “NaNs” to each vector to “stand in” for the values in the other vector that aren’t shared. So, I want them to look like:

A = [1 NaN NaN NaN 5 NaN 20 30 53 NaN 70 92 NaN NaN]
B = [NaN 2 3 4 NaN 16 20 30 NaN 60 NaN NaN 95 100]

Some method by which the vector will have placeholders for the value of the other vector.

Do I combine the vectors, sort it, then somehow search and replace all values from the other vector with NaNs? That seems like a bit of a clunky solution, though not impossible. I feel like there is some more elegant way to accomplish this that I am missing.

Thanks!

Answer 1

Here is one solution using a simple map:

A = [1 5 20 30 53 70 92]
B = [2 3 4 16 20 30 60 95 100]

% map all A and B elements
% use 1 for A and 2 for B
map = zeros(max([A,B]),1);
map(A) = 1;
map(B) = bitor(map(B), 2);

% find the values present in either A, or B
[~,~,j] = find(map);
AA = nan(size(j));
BB = nan(size(j));
AA(bitand(j,1)~=0) = A;
BB(bitand(j,2)~=0) = B;

Comparison with Rodys solution shows this method is a bit faster:

A = unique(randi(10000, 1000, 1));
B = unique(randi(10000, 1000, 1));

tic;
for i=1:1000
    map=zeros(10000,1);
    map(A) = 1;
    map(B) = bitor(map(B), 2);
    [~,~,j] = find(map);
    AA = nan(size(j));
    BB = nan(size(j));
    AA(bitand(j,1)~=0) = A;
    BB(bitand(j,2)~=0) = B;
end
toc


tic
for i=1:1000
    C = union(A,B);
    Ap = NaN(size(C));   
    Ap(ismember(C,A)) = A;
    Bp = NaN(size(C));   
    Bp(ismember(C,B)) = B;
end
toc

isequalwithequalnans(BB, Bp)
isequalwithequalnans(AA, Ap)

Elapsed time is 0.283828 seconds.
Elapsed time is 0.457204 seconds.

ans =
 1

ans =
 1