Hello
I have a vector of 1400x1 with several adjacent values (values with a difference of less than 1).
For example, my data looks like
A = [1 5 6 7 8 9 10 11 12 20 21 22 23]'
I'd like to delete the row that is adjacent to the previous one, so my expected outcome will be [1; 5; 20]
Any ideas on how to do this? I appreciate your help!

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Davide Masiello
Davide Masiello 2022년 10월 25일

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A = [1 5 6 7 8 9 10 11 12 20 21 22 23];
A([false,diff(A) == 1]) = []
A = 1×3
1 5 20

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Nao
Nao 2022년 10월 25일
thank you so much for your swift response.
I get an error with this saying:
Error using horzcat
Dimensions of matrices being concatenated are not consistent
Probably because the data is lined vertically in rows (its a vector of 1400 x 1). I'm very new at this so apologize for my lack of understanding;(
Davide Masiello
Davide Masiello 2022년 10월 25일
편집: Davide Masiello 2022년 10월 25일
Yes, if you have a column vector then you need to write
A([false;diff(A) == 1]) = []
Nao
Nao 2022년 10월 25일
it worked!! thank you so much!!
But since you said "a difference of less than 1" you'd want <1 instead of == 1
A([false;diff(A) < 1]) = []
Of course then your expected outcome is not right. Maybe you mean "a difference of 1 or less" in which case it would be
A([false;diff(A) <= 1]) = []
Nao
Nao 2022년 10월 27일
thats a great point! thank you!

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Nao
2022년 10월 25일

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Nao
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