Matlab accuracy (when 1-1~=0)

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Damian Maxwell 2022-9-20

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https://ww2.mathworks.cn/matlabcentral/answers/1808635-matlab-accuracy-when-1-1-0

评论： Damian Maxwell 2022-9-21

Below please find a screenshot from a sample script that really puzzles me.

When defining simple variables and then substracting them there appear to be tiny error (3e-18) that comes in with surprising results.

It would be great if somebody could explain why this happens and how to avoid it.

w=0.026
b=0.024
i=0.001
w-0.026
b-0.024
i-0.001
w-(b+2*i)
w==(b+2*i)

回答（1 个）

Eric Delgado 2022-9-20

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1808635-matlab-accuracy-when-1-1-0#answer_1057155

编辑：Eric Delgado 2022-9-20

在 MATLAB Online 中打开

It's float operation universe. :)

w=0.026;
b=0.024;
i=0.001;
w-0.026;
b-0.024;
i-0.001;

Instead of:

w == (b+2*i)
ans = logical
   0

Use:

abs(w - (b+2*i)) <= 1e-5 % You could use 1e-17 and still will receive a true logical value
ans = logical
   1

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Walter Roberson 2022-9-20

1/10 is not exactly representable in finite binary floating point -- for the same mathematical reason that 1/3 is not exactly representable in finite decimal.

Suppose we were working in decimal and said 1/3 = 0.3333333333 then if we add those together 3 times we get 0.9999999999 -- which in decimal is distinct from 1.0 . In decimal you need a literally infinite number of digits of precision for the 0.3-repeated added together 3 times to produce a result that is mathematically exactly equal to 1 .

So the problem is not exactly with the fact that MATLAB uses binary: the problem is that if you choose any finite-length fixed-point base, there will always be rational fractions that cannot be exactly represented in finite length. It is an unfortunate mathematical limitation of the Universe.

Damian Maxwell 2022-9-21