# Check values under a certain threshold with an undefined window

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bfenst on 13 Apr 2018
Answered: Image Analyst on 13 Apr 2018
Hello all,
I am new to Matlab. I have a large vector called A (over 400000 values). I would like to find the maximum value of that vector then look for values that are 10 % around the maximum.
Here is an example:
A=[9 10 47 50 49 48 2 46];
mymax=max(A); % in this case mymax=50
mincondition=mymax*0.9; % mincondition=45
Then I should be able to get all values around A(4)=50 that are over mincondtion. The window of considered values stops when mincondition is no longer surpassed. In this case, the results would be:
res=[47 50 49 48];
Thank you.

David Fletcher on 13 Apr 2018
Why isn't the 46 (last element) included in res? (sorry I didn't really understand what you mean by 'the window of considered values stops when mincondition is no longer surpassed' - I am just assuming that you want everything from A that is over mincondition)
bfenst on 13 Apr 2018
Thank you for checking my question. I want everything from A that is over mincondition AND that is consecutive to the max or the values that are over mincondition.
That is why 46 is not included in the res.

David Fletcher on 13 Apr 2018
A=[9 10 47 50 49 48 2 46];
mymax=max(A); % in this case mymax=50
mincondition=mymax*0.9; % mincondition=45
indexer=A>mincondition
[start,stop]=regexp(char(indexer+48),'1+')
res=A(indexer(1:stop(1)))
It's probably worth checking it works thoroughly - I kept getting disturbed by people actually wanting me to do some work while I was writing it.

#### 1 Comment

bfenst on 13 Apr 2018
Thank you very much, it is exactly what I needed. Impressive how a few lines of code do the deed, I was at 30 lines trying to achieve with no results.

Image Analyst on 13 Apr 2018
Here's one way using morphological reconstruction:
A = [9 10 47 50 49 48 2 46];
mask = A >= 0.9 * max(A)
indexes = imreconstruct(A == max(A), mask) % Requires Image Processing Toolbox.
finalValues = A(indexes)