필터 지우기
필터 지우기

Is it ok to have rounding errors when rounding to integers?

조회 수: 2 (최근 30일)
Andreas Brinch Nielsen
Andreas Brinch Nielsen 2017년 1월 19일
편집: dpb 2017년 1월 20일
Run the following:
round(100000,-5)
In my opinion it should return 100000 exactly, but it returns 100000+eps(100000). Do you think this is a bug, or is it an expected round off error because of floating point arithmetic?

채택된 답변

dpb
dpb 2017년 1월 19일
편집: dpb 2017년 1월 20일
Does appear to be a "quality of implementation" issue. Hadn't ever used it w/ the negative optional argument but looks like does have some artifacts, not clear quite where this one would come from. Since it's builtin function can't see implementation but that's what it looks like happens...altho for this input the "deadahead"
>> fix(100000/1E5)*1E5
ans =
100000
>>
works so not sure what must be going on internally.
Looks to me to be worth a bug report.
A workaround I'm sure you've already thought of is to wrap the call in another round...
>> round(round(100000,-5))
ans =
100000
>>

추가 답변 (3개)

John D'Errico
John D'Errico 2017년 1월 19일
편집: John D'Errico 2017년 1월 19일
In general, you cannot represent numbers (that have digits to the right of the decimal point) exactly in floating point arithmetic. So when you use round in this form, you are rounding to the 5th decimal place to the right of the decimal point.
The problem is, only you know that the result should be an integer, because it was an integer before the "round" operation.
round(100000,-5) == 100000
ans =
logical
0
sprintf('%0.55f',round(100000,-5))
ans =
100000.0000000000145519152283668518066406250000000000000000000
What happened is round (with a second operand) works by scaling the number, then performing a round, and then scaling back. But it depends on how the code was written internally, and we don't see the code for round.
This works:
x = 100000;
round(x*10^5)/10^5 == x
ans =
logical
1
But this fails:
round(x/10^-5)*10^-5 == x
ans =
logical
0
My guess is that for negative second argument, this is what they do. However, it would be better to do different operations, depending on whether the second argument is positive or negative.
  댓글 수: 6
Andreas Brinch Nielsen
Andreas Brinch Nielsen 2017년 1월 20일
Floating point numbers do not have problems representing integer numbers. Of course, when you start doing operations on them, these can introduce inaccuracies (as in this example), but at least in this example I would say that it is because of a bad implementation.
dpb
dpb 2017년 1월 20일
"Floating point numbers do not have problems representing integer numbers." up to the length of the number of digits representable by the chosen precision, of course. That's roughly 15 decimal digits for double.

댓글을 달려면 로그인하십시오.


Guillaume
Guillaume 2017년 1월 19일
In my opinion it's a bug that you should report to Mathworks.
It's ironical that round produces a less rounded number than you started with.

Andreas Brinch Nielsen
Andreas Brinch Nielsen 2017년 1월 20일
I wrote a bug report to Mathworks before posting this question, but they started arguing that it was expected behavior, and I just wanted other opinions. Thanks all.
  댓글 수: 1
dpb
dpb 2017년 1월 20일
I'd refer back to the function description in the documentation as to its definition of what is expected output. (If it were my report as I've harped on the issue there is no actual language specification for years, I'd ask what the internal language design document says about it :) )

댓글을 달려면 로그인하십시오.

카테고리

Help CenterFile Exchange에서 Logical에 대해 자세히 알아보기

태그

제품

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by