Logical Indexing With LinSpace Issues
12 vues (au cours des 30 derniers jours)
Afficher commentaires plus anciens
Lucas Graham
le 8 Mar 2023
Commenté : Walter Roberson
le 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 commentaires
Stephen23
le 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:
Réponse acceptée
Star Strider
le 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 commentaires
Star Strider
le 8 Mar 2023
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.
Plus de réponses (1)
Chris
le 8 Mar 2023
Modifié(e) : Chris
le 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 commentaires
Walter Roberson
le 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
Voir également
Catégories
En savoir plus sur Matrix Indexing dans Help Center et File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!