Use array values as assignment indecis
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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;
3 commentaires
Réponse acceptée
Matt J
le 23 Nov 2022
Modifié(e) : Matt J
le 23 Nov 2022
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
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=find(Idcs) %Shouldn't be needed. Use logical indices above.
5 commentaires
Matt J
le 8 Mai 2023
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.
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