Hi all,
This is probably a simple question, but I would love some input from those who can lend it. I have a vector [A] I would like to index according to a condition comparing the nth entry of A with the one following it (n+1).
For instance, if A is [1,2,3,4,5,6,7,6,5,4,3,2,1], and I would like to know which entries of A are larger than the entry preceding it, I would try to do the following:
for i = 1:size(A)
if i > 1
j = A(i) > A(i-1);
end
end
Where j = [0,1,1,1,1,1,1,0,0,0,0,0,0]
Is there a way to do this without using a loop? Any help would be appreciated, thank you.

 Réponse acceptée

Renato Agurto
Renato Agurto le 19 Avr 2016
j = [0 A(2:end) > A(1:end-1)]

2 commentaires

Tyler
Tyler le 19 Avr 2016
Awesome, thank you so much! If I wanted to do something more complicated, such as wanting the absolute value of the difference of the two entries to be greater than a certain value, could I do that?
Aa = [0 abs(A(1:end)-A(1:end-1)) > .001]
This doesn't quite work, but I might be forgetting something obvious.
Walter Roberson
Walter Roberson le 19 Avr 2016
You used A(1:end)-A(1:end-1) which subtracts two vectors of different sizes. Renato used A(2:end) for the first operation.

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Plus de réponses (1)

Walter Roberson
Walter Roberson le 19 Avr 2016
[abs(diff(A)) > 0.001, false]
The false compensates for the fact that diff(A) is one shorter than A itself. You will find that in practice it is not a problem to use a logical index shorter than the array dimension, so for most purposes
abs(diff(A)) > 0.001
would be just as good.

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