I am facing a problem with ismember subfunciton in Matlab.

I am facing a problem with ismember subfunciton in Matlab.

UseCase:

 Simulation_Time_Vecort = 0:0.01:42;

 TestVector = 

     0    1.0000
    0.7000         0
    0.8000    1.0000
    7.3100         0
    7.4100    1.0000
   13.6200    1.0000
   20.3200    1.0000
   27.0200    1.0000
   33.0200    1.0000
   39.9200    1.0000

Simulation_Values_Vecort = zeros(1,length(Simulation_Time_Vecort));
FlagSignalName_Value = [Simulation_Values_Vecort ; Simulation_Values_Vecort]';
FlagSignalName_Value(:,1) = double(Simulation_Time_Vecort);
FlagSignalName_Value(:,2) = ismember(Simulation_Time_Vecort,TestVector_Data(:,1)');

Problem: When I try to excute this script, ismember is not executed as expected. For some cases though the value exist in both vectors used in ismember function states above, output is not displayed or updated as value 1. In my case for value ‘0.7000’ ismember result is displayed as 0

Answer 1

[Walter Roberson]

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

If you have a new enough MATLAB version, use ismembertol

[Guillaume]

"If you have a new enough MATLAB version, use ismembertol"

Alternatively, round all values to 2 decimals, since the data shown does not appear to have more precision

FlagSignalName_Value(:,2) = ismember(round(Simulation_Time_Vecort, 2), round(TestVector_Data(:,1)', 2));

edit: as pointed out by Walter, due to the way floating point works, there's no guarantee that

round(somevaluenearly0.7, x)

and

round(someothervaluenearly0.7, x)

will produce the exact same result. With x = 2, I'm hoping it shouldn't be the case (but I haven't tested it)

[Walter Roberson]

Ah, but the rounding could come out differently.

[Guillaume]

To two decimal place? Hopefully not. Otherwise, I'd say that round is badly implemented.

[Stephen23]

Surely a tolerance is the only robust solution. For example (I used this round2sf):

>> A = 0.996;
>> B = 0.994;
>> round2sf(A,2)
ans =  1.0000
>> round2sf(B,2)
ans =  0.99000

The values differ by only 0.002, but rounding to 2 sigfig will make them diverge, not converge. Using a tolerance would resolve this. All pairs of values that differ by a little less than least significant figure and that are on the rounding boundary will diverge...

[Guillaume]

Ah yes, but your values differ on the 3rd significant digit. From the example data provided, I'm assuming that the values only differ by a few eps.

Yes, tolerance comparison is more robust. My suggestion was for if you don't have ismembertol.