Logical Indexing With LinSpace Issues

Question

Lucas Graham 2023-3-8

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1924875-logical-indexing-with-linspace-issues

评论： Walter Roberson 2023-3-9

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))
    1.3000
flag = (X == 1.3);
any(flag) %Yet for some reason 1.3 isn't detected.
ans = logical
   0
%Just to be absolutely sure I'm not mad
disp(X(12))
    1.3000
X(12) == 1.3
ans = logical
   0
% Yet other non-integers still work! 
flag = (X == 1.1);
any(flag)
ans = logical
   1

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Stephen23 2023-3-8

"Any ideas or help or explanation would be appreciated."

This is a completely expected result with binary floating point numbers:

https://www.mathworks.com/matlabcentral/answers/57444-faq-why-is-0-3-0-2-0-1-not-equal-to-zero

https://www.mathworks.com/matlabcentral/answers/316889-why-two-equal-numbers-are-not-equal

http://www.mathworks.com/matlabcentral/answers/102419-how-do-i-determine-if-the-error-in-my-answer-is-the-result-of-round-off-error-or-a-bug

http://www.mathworks.com/matlabcentral/answers/140656-about-numerical-difficulty-of-equality-comparison-operation-between-two-double-real-numbers-in-matlab

This is worth reading as well:

https://docs.oracle.com/cd/E19957-01/806-3568/ncg_goldberg.html

http://faculty.tarleton.edu/agapie/documents/cs_343_arch/papers/1991_Goldberg_FloatingPoint.pdf

Lucas Graham 2023-3-8

Thank you for these resources!

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Answer 1

Star Strider 2023-3-8

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1924875-logical-indexing-with-linspace-issues#answer_1187965

在 MATLAB Online 中打开

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. 
X = 1×29
   0.200000000000000   0.300000000000000   0.400000000000000   0.500000000000000   0.600000000000000   0.700000000000000   0.800000000000000   0.900000000000000   1.000000000000000   1.100000000000000   1.200000000000000   1.300000000000000   1.400000000000000   1.500000000000000   1.600000000000000   1.700000000000000   1.800000000000000   1.900000000000000   2.000000000000000   2.100000000000000   2.200000000000000   2.300000000000000   2.400000000000000   2.500000000000000   2.600000000000000   2.700000000000000   2.800000000000000   2.900000000000000   3.000000000000000
disp(X(12))
   1.300000000000000
flag = (X == 1.3);
any(flag) %Yet for some reason 1.3 isn't detected.
ans = logical
   0
Check_Equality = 1.3 - X(12)
Check_Equality = 
     2.220446049250313e-16
%Just to be absolutely sure I'm not mad
disp(X(12))
   1.300000000000000
X(12) == 1.3
ans = logical
   0
% Yet other non-integers still work! 
flag = (X == 1.1);
any(flag)
ans = logical
   1

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.

See Floating-Point Numbers and Accuracy of Floating-Point Data for an extensive discussion.

.

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Lucas Graham 2023-3-8

Thank you!

Star Strider 2023-3-8

My pleasure!

If my Answer helped you solve your problem, please Accept it!

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Answer 2

Chris 2023-3-8

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1924875-logical-indexing-with-linspace-issues#answer_1187970

编辑：Chris 2023-3-8

在 MATLAB Online 中打开

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))
ans = 
    '1.199999999999999955591079014993738383054733276367187500000000
     1.199999999999999955591079014993738383054733276367187500000000'
% X(12) doesn't equal the double for 1.3.
sprintf('%.60f\n%.60f',X(12),Y(12))
ans = 
    '1.299999999999999822364316059974953532218933105468750000000000
     1.300000000000000044408920985006261616945266723632812500000000'

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.

https://www.geeksforgeeks.org/ieee-standard-754-floating-point-numbers/

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Lucas Graham 2023-3-8

编辑：Lucas Graham 2023-3-8

My intuition is that using MATLAB's colon operator to build arrays is less likely to run into float point errors because it probably uses addition instead of division, is that correct?

For example, using like you did 0.2:0.1:3 vs. linspace(0.2,3,29).

Walter Roberson 2023-3-9

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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