Why does MATLAB not finding a element from my matrix?

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x = 0.4:0.01:0.5
x = 1x11
0.4000 0.4100 0.4200 0.4300 0.4400 0.4500 0.4600 0.4700 0.4800 0.4900 0.5000
idx = find(x==0.41)
idx = 1x0 empty double row vector
  3 Comments
Tshikaya
Tshikaya on 27 Mar 2024
Thank you for the tip. My problem was actually the following:
A = zeros(1,100);
x = zeros(1,100);
for i = 1:100
x(i) = 0.01*randi([38 45],1,1);
if ismember(0.41,x(i));
A(i) = 10;
else
A(i) = -1;
end
end
So I was shoked to see that x was sometimes 0.41 but still not picking from the code. I have just changed ismember by ismembertol and now everything is fine.
Walter Roberson
Walter Roberson on 24 Jun 2025
Note that the colon operator is handled as repeated addition, so 0.4, 0.4+0.1, 0.4+0.1+0.1, 0.4+0.1+0.1+0.1 and so on. Because double precision numbers operate in binary instead of in decimal, 0.1 cannot be exactly represented, and instead each 0.1 addition adds 0.1000000000000000055511151231257827021181583404541015625
Mathworks could have chosen an impletation of (initial value) + increment * (position in the sequence minus 1), so 0.4 + [0, 1, 2, 3, 4 ...] * 0.1 ... but Mathworks did not chose that (possibly for efficiency reasons.)

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Accepted Answer

Stephen23
Stephen23 on 27 Mar 2024
Moved: Dyuman Joshi on 27 Mar 2024
Testing for exact equivalence of binary floating point numbers should be avoided.
Compare the values, are they actually the same? (hint: no):
x = 0.4:0.01:0.5;
fprintf('%.40f\n', x(2), 0.41)
0.4100000000000000310862446895043831318617 0.4099999999999999755750934582465561106801
The recommended approach is to compare the absolute difference against a tolerance:
tol = 1e-5;
abs(x(2)-0.41)<tol
ans = logical
1
More information on this topic:
This is worth reading as well:
  2 Comments
Steven Lord
Steven Lord on 24 Jun 2025
If you're using release R2024b or later, you could use the isapprox function to determine if two numbers are approximately equal.

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