How to pick next value from vectors?

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Teemu
Teemu on 4 Mar 2013
Edited: shariq khan on 3 Dec 2018
I have a cumulative vector, size is about 200. I want to pick up that index where value is 0,5 or value which is next to it. For example (0.1, 0.23, 0.42, 0.49, 0.52, 0.56,...), so i want to pick up index which corresponding to value 0.52. How can I do it?
  3 Comments
Teemu
Teemu on 5 Mar 2013
Minimal distance, in this case it will be 0.49
Teemu
Teemu on 5 Mar 2013
actually I need both

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Accepted Answer

Jos (10584)
Jos (10584) on 4 Mar 2013
Here is one approach:
v = [0.1 0.23 0.4 0.52 0.56 1.1]
idx = find(v >= 0.5,1,'first')
v(idx)
  8 Comments
Jan
Jan on 5 Mar 2013
Please, Teemu, read my answer below. There you find the code to pick the element with the minimal distance. It is the 2nd line and the index is the 2nd output of min(abs(v - 0.5)).
Teemu
Teemu on 5 Mar 2013
Sorry I didn't noticed that, I thought it was answer for next index. Thanks for your patience.

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More Answers (2)

Jos (10584)
Jos (10584) on 5 Mar 2013
To do this for multiple values, take a look at my function NEARESTPOINT:

Jan
Jan on 4 Mar 2013
Edited: Jan on 5 Mar 2013
v = [0.1 0.23 0.4 0.52 0.56 1.1];
[absdist, nearest] = min(abs(v - 0.5)); % **SOLUTION IS HERE** !!!
if absdist == 0 % or: <= eps(v(index))
following = index;
else
following = index + 1;
end
nearestValue = value(nearest);
followingValue = value(following);
  2 Comments
shariq khan
shariq khan on 3 Dec 2018
what if I need next larger value? consider same data but my value at specific index is nil (or no value from dataset) but now I want to find if there is any value next to that specific value in the dataset
shariq khan
shariq khan on 3 Dec 2018
Edited: shariq khan on 3 Dec 2018
I think I found answer to my problem
Problem - all vectors are of same length
x = [some values]
t = [some values]
x1 = 0.9*max(x);
find t value that corresponds to x1 value
if not find the closest value
solution : A lot from @jan ans
[answer,index] = min(abs(x - x1))
t1(that is to be determined) = t(index);
I hope it helps to solve people facing specific problem

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