Violation of logical indexing criteria problem

Hi everyone. I am collecting grid points (alpha) from a larger grid (x) by the following commands:

I = abs(x)<0.5;
alpha = x(I);

My problem is that this criteria is not being able to filter x = -0.5. My question is, does Matlab think abs(-0.5)>0.5?

What am i doing wrong here? Please help.

[Stephen23]

You should read about the precision limits of floating point values:

http://www.mathworks.com/help/matlab/matlab_prog/floating-point-numbers.html

http://www.mathworks.com/matlabcentral/answers/69-why-does-1-2-3-1-3-not-equal-zero

http://matlab.wikia.com/wiki/FAQ#Why_is_0.3_-_0.2_-_0.1_.28or_similar.29_not_equal_to_zero.3F

http://www.mathworks.com/matlabcentral/answers/57444-faq-why-is-0-3-0-2-0-1-not-equal-to-zero

http://www.mathworks.com/company/newsletters/news_notes/pdf/Fall96Cleve.pdf

http://www.mathworks.com/matlabcentral/answers/41536-why-is-2-24-10-22-4-not-equal-to-0

https://www.mathworks.com/matlabcentral/newsreader/view_thread/66553

http://blogs.mathworks.com/loren/2006/08/23/a-glimpse-into-floating-point-accuracy/

http://www.mathworks.com/help/fixedpoint/ug/floating-point-numbers.html

And some external links on this topic:

http://www.lahey.com/float.htm

http://matlabgeeks.com/tips-tutorials/floating-point-comparisons-in-matlab/

Answer 1

[Walter Roberson]

Your value is probably not being calculated as exactly -0.5 . -0.5 is exactly representable in binary floating point, but different ways of calculating what "should" be -0.5 do not always give that exact value. If you add 0.5 to the value you think should be -0.5 you will probably find that the result is not exactly 0.

[Sudipta Ray]

Actually yes, my -0.5 value is getting calculated as

 x= cos(2*pi/3).

I want to know if there is a way to check the inclusion of the number within a tolerance value, like the command

ismembertol(xn,x)

[Walter Roberson]

-0.5 - tol < x & x < 0.5 + tol

There are some inrange() contributions in the File Exchange. You need to watch out for the boundary conditions for them.

[Steven Lord]

 x = cos(2*pi/3);
 y = -0.5;
 format hex
 [x; y]

The value stored in x is not exactly -0.5. It's close, but you can see that the bit patterns of the numbers aren't the same.

 format
 difference = x-y

Why is this, you may ask? The PI function doesn't return the irrational value representing the exact ratio between the circumference of a circle and its diameter, for one thing. Instead, PI returns the double precision value closest to the irrational value. That's not enough to compute the circumference of the universe to the nearest atom (that would require about 39 digits according to the Wikipedia page for pi) but it's good enough for most computations.

Answer 2

[Image Analyst]

If you want =, then use =

indexesInRange = abs(x) <= 0.5;
alpha = x(indexesInRange);

[Sudipta Ray]

Sorry about the ambiguity. I want to filter "out" x = -0.5, not include it. That is why I used "strictly less than (<)" condition. Matlab is somehow finding

abs(-0.5)< 0.5

and the value x = -0.5 is getting into my selection, which I don't want.

Answer 3

[Sudipta Ray]

I have tried some fix, now it is working ok

tol = 1e-15; 
I = abs(x)<r-tol;

still, if anyone has a better solution, please post it. I have considered "r-tol" because that solves my problem and not "r+tol". But this is too specific and works because the value

abs(cos(2*pi/3)) > 0.5.

That is why for my case only x = -0.5 was creating a problem, and not x = 0.5.