Logical Indexing With LinSpace Issues
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Logical indexing is missing an equivalence in a linspace array. I have absolutely no idea why or how to fix this. I'm running MATALB R2022b on Windows 11.
Any ideas or help or explanation would be appreciated.
Here is how you can recreate my issue:
X = linspace(0.2,3,29); %Creates an array going from 0.2 -> 3 via steps of 0.1. So 1.3 is in it.
disp(X(12))
flag = (X == 1.3);
any(flag) %Yet for some reason 1.3 isn't detected.
%Just to be absolutely sure I'm not mad
disp(X(12))
X(12) == 1.3
% Yet other non-integers still work!
flag = (X == 1.1);
any(flag)
2 comentarios
Stephen23
el 8 de Mzo. de 2023
"Any ideas or help or explanation would be appreciated."
This is a completely expected result with binary floating point numbers:
This is worth reading as well:
Lucas Graham
el 8 de Mzo. de 2023
Respuesta aceptada
Más respuestas (1)
This looks like a problem of computer precision. Double-precision floating point numbers in matlab (and floating-point numbers in general) a̶r̶e̶ ̶n̶o̶t̶ ̶e̶x̶a̶c̶t̶ (edit: are a finite set of discrete values, so they are usually unlikely to exactly match the value we see displayed)
X = linspace(0.2,3,29);
Y = 0.2:0.1:3;
% X(11)==1.2 because the result of the linspace division
% is the same as the double representation for 1.2
sprintf('%.60f\n%.60f',X(11),Y(11))
% X(12) doesn't equal the double for 1.3.
sprintf('%.60f\n%.60f',X(12),Y(12))
You can use other comparison operators to test rough equivalence, e.g., (X(12)-1.3) < 1e-14
Here's a bit more information, if you're interested.
4 comentarios
Lucas Graham
el 8 de Mzo. de 2023
Walter Roberson
el 8 de Mzo. de 2023
Well, in a sense, double precision is exact... but only for a subset of rationals, some of the rationals that have powers of two as the denominator (not all of those). It is however not exact for powers of 10 in the denominator (just like finite decimal is not exact for powers of 3 or 7 in the denominator)
Lucas Graham
el 8 de Mzo. de 2023
Editada: Lucas Graham
el 8 de Mzo. de 2023
Walter Roberson
el 9 de Mzo. de 2023
No, the opposite. Each time you add two floating point numbers, the round-off error can increase (unless the sequence of operations has been carefully chosen.) The error for 0.1, 0.1+0.1, 0.1+0.1+0.1 is greater than for (1:3)/10
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