Logical Indexing With LinSpace Issues
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Lucas Graham
il 8 Mar 2023
Commentato: Walter Roberson
il 9 Mar 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 Commenti
Stephen23
il 8 Mar 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:
Risposta accettata
Star Strider
il 8 Mar 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 Commenti
Star Strider
il 8 Mar 2023
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.
Più risposte (1)
Chris
il 8 Mar 2023
Modificato: Chris
il 8 Mar 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 Commenti
Walter Roberson
il 9 Mar 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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