If value in one vector is in another, assign the value of the other column of the second vector to a third

I have 3 vectors (these are simplified versions I created as the real data points are vast and I don't know how to put them in here and I need the concept of the task, but this means I can't solve this by assigning specific values I need a general if statement).

T = 0:20;
s = zeros(1,lenth(T));
stimuli = [4 8 14 2 16 6 15 12 3 ; 6 8 15 12 3 11 19 16 2];

I want to see if values of T are in row 2 of stimuli, and if they are I want to assign the corresponding value in row 1 of stimuli to the column in s equal to time T.

Eg, if T=19, which is in row 2 of stimuli, I want 15 to become the value in column 20 of s (20 not 19 as I'm including the 0 as the first value)

Here is my current code:

for i = 1:length(T)
if ismember([T(i)],stimuli(2,:)) %if there is a stimulus at time
        s(i) = stimuli(1,(stimuli(2,T(i)))); %assigne the stimulus value at that time to s
    end
end

does anyone know how I can do this? I can provide more info on my project if needed, but hopefully this will be enough :)

Answer 1

[Jan]

I do not understand exactly, what you are asking for. A guess:

T = 0:20;
s = zeros(1, numel(T));
stimuli = [4 8 14 2 16 6 15 12 3 ; 6 8 15 12 3 11 19 16 2];
[match, index] = ismember(T, stimuli(2, :));
s(match) = stimuli(1, index(match))

s = 1×21

0 0 3 16 0 0 4 0 8 0 0 6 2 0 0 14 12 0 0 15 0

Answer 2

[Awais Saeed]

T = 0:20;

0   1   2   3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18  19  20

s = zeros(1,length(T));

0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0  0

stimuli = [4 8 14 2 16 6 15 12 3 ; 6 8 15 12 3 11 19 16 2]

stimuli = 2×9

4 8 14 2 16 6 15 12 3 6 8 15 12 3 11 19 16 2

for ii = 1:1:numel(T)
    % find if T(ii) is in row 2 of stimuli
	idx = find(T(ii) == stimuli(2,:));
	if(isempty(idx) == 0)
        % if yes, pick corresponding column value from row1 in stimuli
        % and put it in s
		s(ii) = stimuli(1,idx);
	end
end
disp(int2str(s))

0   0   3  16   0   0   4   0   8   0   0   6   2   0   0  14  12   0   0  15   0

[Jan]

You can omit the find() and use the faster logical indexing:

	idx = (T(ii) == stimuli(2,:));
	if any(idx)
		s(ii) = stimuli(1,idx);
	end

[Awais Saeed]

Yes, you are right, that is much better.

[Catherine Law]

thanks so much for your help, but this gives me a vector for s whose length is less than T, and I need them the same size for plotting etc later in my code, any idea why this might be happening? I tried making it

s(i+1) = stimuli(1,idx);

as I thought this would also shift them across 1 to account for T=0 being in column 1, but that didn't solve the length issue?

[Catherine Law]

actually nevermind it was an issue elsewhere in my code thanks so so much!!