- are the input values guaranteed to be integers?
- are the duplicate values guaranteed to be next to each other?
- are you guaranteed to have 9 or few duplicates of a given number?
How to change consecutive duplicate values, so that they are unique
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Jasmine Zhu
am 15 Feb. 2023
Kommentiert: Jasmine Zhu
am 15 Feb. 2023
I have an array A = [1,2,2,4,5,5,5,6], and I would like to get B=[1,2,2.1,4,5,5.1,5.2,6]. How to do it? Thanks!
2 Kommentare
the cyclist
am 15 Feb. 2023
A few questions:
Akzeptierte Antwort
the cyclist
am 15 Feb. 2023
Bearbeitet: the cyclist
am 15 Feb. 2023
If the answer to all of my question in my above comments are "yes", then
% Input
A = [1,2,2,4,5,5,5,6];
% Algorithm
B = A;
for ib = 2:numel(B)
if B(ib)==floor(B(ib-1))
B(ib) = B(ib-1) + 0.1;
end
end
B
2 Kommentare
the cyclist
am 15 Feb. 2023
tic
% Input
A = [1,2,2,4,5,5,5,6];
A = repmat(A,1,ceil(2.6e6/numel(A)));
% Algorithm
B = A;
for ib = 2:numel(B)
if B(ib)==floor(B(ib-1))
B(ib) = B(ib-1) + 0.1;
end
end
B;
toc
61 milliseconds in this one test.
Weitere Antworten (1)
John D'Errico
am 15 Feb. 2023
Bearbeitet: John D'Errico
am 15 Feb. 2023
I once wrote an unrounding tool, that did something like what you want to do. The goal I chose was to perform a minimal perturbation to the original squence that was consistent with having rounded the initial vector, yet is still monotonic, AND is as smooth as possible.
However, you need to consider if your goal is fully valid. It seems you want to add 0.1 to each replicated element. But how would your plan work for the vector
V = [1 1,repmat(2,1,15),3 3 3 3 3 3 3]
The thing is, in the vector V, adding multiples of 0.1 will fail.
Vhat = unround(V)
plot([V;Vhat]','o-')
No perturbation to V is greater than 0.5.
norm(V - round(Vhat))
You could do something different of course.
Edit: Of curse, since now I know that your vector has 3 million elements in it, I would expect that a tool that calls quadprog will fail anyway. For the future, it would have been good of you to tell us pertinent information like that, as I would not have bothered to answer with this solution.
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