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Jamie

Identify an array of arbitrary minimum values. Alternatives to min / find ?

Asked by Jamie
on 18 Apr 2013

I would like to identify the occurences (index) of an arbitrary value within a Vector. The values I wish to identify will be close to but may not be exactly equal to this value.

For example having obtained the vector [Vector,t] = lsim(G,u,t) I would like to identify those points where the Vector assumedly intersects a horizontal line of given amplitude plot([X1 X2],[Y1 Y1])

I can obtain the first value quite simply with something along the lines of

[index value] = min(abs(abs(Vector) - abs(arbitrary_value)))

The find function is of some assitance however it requires that the arbitrary value equals the value of an element within the Vector exactly. Hence Idxs = find(A==MinValue) does not return a value within proximity of the intersection (or any point for that matter)

I thought I might be able to scrape by with a for loop i.e.

for i = 1:length(Vector)

if (Vector(i) - arbitrary_value) < tolerance
  index(k) = i;
  Value(k) = y(i);
  k = k+1;
end
end

however it is far from an attractive or precise solution.

Any suggestions would be welcome.

Regards

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1 Answer

Answer by Andrei Bobrov
on 18 Apr 2013
Edited by Andrei Bobrov
on 18 Apr 2013
 Accepted answer

eg:

tolerance = 1e-3;
ii = abs(Vector - arbitrary_value) < tolerance;
Value = Vector(ii);
index = find(ii);

ADD variant

t = vin - av;
idx = cellfun(@(x)strfind(sign(t(:).'),x),{0,[-1 1],[1 -1]},'un',0);
x0 = bsxfun(@plus,[idx{2:end}],(0:1)');
[~,ii] = min(abs(t(x0)));
iout = sort([[idx{1}],x0(sub2ind(size(x0),ii,1:size(x0,2)))]);
index = vin(iout);

  3 Comments

Jamie
on 18 Apr 2013

Hi Andrei, thank you for that. much tidier. With regards to the respective points of intersection, depending on the value of 'tolerance' I either obtain numerous points around certain crossings or fail to obtain other crossings altogeather. If I could obtain one index per point of intersection that would be most helpful

Andrei Bobrov
on 18 Apr 2013

See ADD part in my answer.

Jamie
on 19 Apr 2013

Superb. Many thanks for your assitance


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