equality operator between matrix and scalar

1 Ansicht (letzte 30 Tage)
A
A am 17 Apr. 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.

Akzeptierte Antwort

Matt Kindig
Matt Kindig am 17 Apr. 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

Weitere Antworten (3)

Kye Taylor
Kye Taylor am 17 Apr. 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 am 17 Apr. 2012

James Tursa
James Tursa am 17 Apr. 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:

Kategorien

Mehr zu Special Functions finden Sie in Help Center und File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by