Logical Indexing With LinSpace Issues
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Lucas Graham
el 8 de Mzo. de 2023
Comentada: Walter Roberson
el 9 de Mzo. de 2023
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:
Respuesta aceptada
Star Strider
el 8 de Mzo. de 2023
format long
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.
Check_Equality = 1.3 - X(12)
%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)
The‘Check_Equality’ assignment demonstrates that the two numbers are actually not equal.
You’re not mad! You just haven’t been introduced to the subtle mysteries of floating-point numeric repreesentation.
.
2 comentarios
Star Strider
el 8 de Mzo. de 2023
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.
Más respuestas (1)
Chris
el 8 de Mzo. de 2023
Editada: Chris
el 8 de Mzo. de 2023
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
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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