Issue with find() function

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Mitchell Frankel
Mitchell Frankel on 5 Jun 2014
Answered: Image Analyst on 5 Jun 2014
If I create an array x = 0.1:0.05:0.6; and then ask find(x==0.15), I get an empty matrix. It also fails for x==0.40? I don't understand this at all.
  1 Comment
Cedric
Cedric on 5 Jun 2014
If you understand the answers below, read about EPS. In the end, you will probably end up doing something like:
pos = find( abs(x - 0.15) < eps(1) ) ;

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Accepted Answer

Star Strider
Star Strider on 5 Jun 2014
It’s called ‘floating point approximation error’. The concept is similar to the decimal approximation of 1/3 = 0.33333.... To get around it, you need to state the conditions in your find calls to allow for some imprecision.
Run these to get an idea of how the concept applies to your problem:
x = 0.1:0.05:0.6;
ix11 = find(x <= 0.15, 1, 'last')
ix12 = find(x >= 0.15, 1, 'first')
x1 = x([ix11 ix12])
ix21 = find(x <= 0.40, 1, 'last')
ix22 = find(x >= 0.40, 1, 'first')
x2 = x([ix21 ix22])

More Answers (2)

Azzi Abdelmalek
Azzi Abdelmalek on 5 Jun 2014

Image Analyst
Image Analyst on 5 Jun 2014
Here's some pretty general code based on the FAQ Azzi referred you to:
x = 0.1 : 0.05 : 0.6 % Sample data
targetValue = 0.15 % Whatever value you want to find
tolerance = .005 % We want to find x within this value of the target.
% Find values in range.
indexesInRange = abs(x - targetValue) <= tolerance
xThatAreInRange = x(indexesInRange)
% Or, in an if statement, if you want to use it in an if statement instead...
if abs(x(2) - targetValue) <= tolerance
% x(2) is close enough to the targetValue.
else
% x(2) is far away from the targetValue.
end

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