Use array values as assignment indecis

A = N-by-2 array, wherein A(:,1)<A(:,2)

B = M-by-2 array, wherein B(:,1)<B(:,2)

A and B are essentially time stamps.

C = unique(vertcat(A,B));

[~,D] = ismember(A,C);

[~,E] = ismember(B,C);

C([D(1,1):D(1,2),...D(N,1):D(N,2)],2) = any scalar, or an array with a sufficient number of elements to dstribute.

C([E(1,1):E(1,2),...E(N,1):E(N,2)],3) = any scalar, or an array with a sufficient number of elements to dstribute.

I'm hung up on defining those assignment indecis [D(1,1):D(1,2),...D(N,1):D(N,2)] without using a loop, i.e.

IdcsD = double.empty;
for i=1:size(D,1)
    Idcs = [Idcs,[D(i,1):D(i,2)]];
end
C(IdcsD) = Value;

[Matt J]

What is the size of D and what is the size of 'Value'?

[Matt J]

Please show an example for small N and showing the desired output D.

[Gabriel Stanley]

Value is a scalar, while OBE

Answer 1

[Matt J]

The question has nothing to do really with ismember. It's more about how to turn a list of interval end points into indices. One way is as follows:

D=[1 5 9;
   2 7 12]'  %interval end points

D = 3×2

1 2 5 7 9 12

C=rand(1,12);  

nc=length(C);

assert(all(diff(D,1,2)>=0)   & all(D(:)>0)  & all(D(:)<=nc) ,'Invalid data present')


D=D+[0,1];  s=0*D+[1,-1];

csum=cumsum( accumarray(D(:),s(:),[nc+1,1]) );

Idcs=logical(csum(1:end-1))',

Idcs = 1×12 logical array
   1   1   0   0   1   1   1   0   1   1   1   1

Idcs=find(Idcs) %Shouldn't be needed. Use logical indices above.

Idcs = 1×9

1 2 5 6 7 9 10 11 12

[Gabriel Stanley]

I didn't think that my initial (or revised) question implied that I was asking about ismember functionality, only that I was using ismember in my process. That said, your answer is exactly what I was looking for. Thank you.

[Gabriel Stanley]

Follow-up coming back to what seems to be an improved version of this: why have the

C = rand(1,12);
nc = length(C);

instead of just

nD = max(D,[],'all');

?

[Matt J]

I see no reason to assume nC and nD will be related in general, even though they were in the above example.

[Gabriel Stanley]

So in the case of the interval end points of D defining a subset of some larger data set C, nC should be set to the maximum index of the relevant dimension in C?

[Matt J]

The dimension/shape of C seems irrelevant to this task. I think you would just flatten C to C(:) and apply the code above as is.