Finding array index for a given array element does not work all the time

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Hello,
I've been using the find(ismemeber()) function to find the index of an array element, however it does not work all the time for some reason. As you can see from the picture, it returns and empty vector at some given search values. As it can be seen from the array, it should return 393 for the last line of code. I am using the latest installation of matlab at the time of writting (2023b update 7).
To me this looks like a bug with the ismemeber() function, as it returns a false for the second entry(0.3920) but not the first entry(0.3910), se last two lines of code below. However, it could just be that i am doing something wrong? As I construct the array with a start, end, and stepsize, it shouldn't be a round of error?
Best regards Anders
mf = 0:0.001:0.6;
find(ismember(mf,0.3910))
ans = 392
find(ismember(mf,0.3920))
ans = 1×0 empty double row vector
any(ismember(mf,0.3910))
ans = logical
1
any(ismember(mf,0.3920))
ans = logical
0
  1 Kommentar
Stephen23
Stephen23 am 1 Mär. 2024
"To me this looks like a bug with the ismemeber() function"
Nope.
"However, it could just be that i am doing something wrong?"
Yes. You are performing different operations on some binary floating point numbers and are then incorrectly assuming that these will produce exactly the same values. However that is not the case.
This is a completely expected result with binary floating point numbers:
This is worth reading as well:

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Akzeptierte Antwort

Dyuman Joshi
Dyuman Joshi am 1 Mär. 2024
Welcome to the world of floating point numbers, where not all numbers can be represented exactly in binary form.
When comparing floating point numbers, the best practice is to use a tolerance -
mf = 0:0.001:0.6;
val = 0.3910;
tol = 1e-6;
idx1 = find(abs(mf - val) < tol)
idx1 = 392
You can also use ismembertol here -
out = ismembertol(mf,val);
idx2 = find(out)
idx2 = 392

Weitere Antworten (1)

Chunru
Chunru am 1 Mär. 2024
Mind the round off error in searching double array
mf = 0:0.001:0.6;
i = find(abs(mf - 0.3910) < 1e-6)
i = 392

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