Use array values as assignment indecis

2022 11 23
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更新时间：2023 5 8
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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;

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Matt J
Matt J 2022-11-23
What is the size of D and what is the size of 'Value'?
Matt J
Matt J 2022-11-23
Please show an example for small N and showing the desired output D.
Gabriel Stanley
Gabriel Stanley 2022-11-23
编辑：Gabriel Stanley 2022-11-23
Value is a scalar, while OBE

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 采纳的回答

Matt J
Matt J 2022-11-23
编辑：Matt J 2022-11-23
Ran in:

0 个投票

Ran in:
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

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Gabriel Stanley
Gabriel Stanley 2022-11-23
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
Gabriel Stanley 2023-5-8
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
Matt J 2023-5-8
I see no reason to assume nC and nD will be related in general, even though they were in the above example.
Gabriel Stanley
Gabriel Stanley 2023-5-8
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
Matt J 2023-5-8
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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