equality operator between matrix and scalar

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A
A el 17 de Abr. de 2012
consider:
a=[-1:0.1:1];
c=a==.1;
it returns c as a matrix of nulls; while I expect c(12) to be 1.

Respuesta aceptada

Matt Kindig
Matt Kindig el 17 de Abr. de 2012
This is due to floating point precision errors, explained here: http://blogs.mathworks.com/loren/2006/08/23/a-glimpse-into-floating-point-accuracy/
A better way to do this is by comparing against a tolerance, such as:
c = abs(a-0.1)<=eps

Más respuestas (3)

Kye Taylor
Kye Taylor el 17 de Abr. de 2012
The value .1 and a(12) differ by only one bit as a result of numerical round-off. You can see this with the commands
num2hex(.1)
num2hex(a(12))
A better way to test equality takes into account the possibility of numerical round-off. For example, create your new c variable with the command
c = abs(a-0.1)<=eps(max(a));

Walter Roberson
Walter Roberson el 17 de Abr. de 2012

James Tursa
James Tursa el 17 de Abr. de 2012
You can try this to see what the numbers are exactly:
num2strexact((-1:0.1:1)')
Sometimes a beter way to do this is to have an array with integer values to start with and then create a 2nd array with the fractional values. E.g.,
A = -10:10;
a = A / 10;
c = A==1;
You can use a for downstream calculations just like before, and you can use c for the value testing.
You can find num2strexact on the FEX here:

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