Logical Indexing With LinSpace Issues
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Lucas Graham
on 8 Mar 2023
Commented: Walter Roberson
on 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 Comments
Stephen23
on 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:
Accepted Answer
Star Strider
on 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 Comments
Star Strider
on 8 Mar 2023
My pleasure!
If my Answer helped you solve your problem, please Accept it!
.
More Answers (1)
Chris
on 8 Mar 2023
Edited: Chris
on 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 Comments
Walter Roberson
on 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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