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equality operator between matrix and scalar

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A
A 2012-4-17
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.

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采纳的回答

Matt Kindig
Matt Kindig 2012-4-17
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

更多回答（3 个）

Kye Taylor
Kye Taylor 2012-4-17
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 2012-4-17
James Tursa
James Tursa 2012-4-17
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:
http://www.mathworks.com/matlabcentral/fileexchange/22239-num2strexact-exact-version-of-num2str

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