[R] why the base::round(0.015, 2) returns 0.02?

[Richard M. Heiberger]

interesting.  this looks like an OS problem, since ?round says
     ‘round’ rounds the values in its first argument to the specified
     number of decimal places (default 0).  See ‘Details’ about “round
     to even” when rounding off a 5.
Details
     Note that for rounding off a 5, the IEC 60559 standard (see also
     ‘IEEE 754’) is expected to be used, ‘_go to the even digit_’.
     Therefore ‘round(0.5)’ is ‘0’ and ‘round(-1.5)’ is ‘-2’.  However,
     this is dependent on OS services and on representation error
     (since e.g. ‘0.15’ is not represented exactly, the rounding rule
     applies to the represented number and not to the printed number,
     and so ‘round(0.15, 1)’ could be either ‘0.1’ or ‘0.2’).

.015 is right on the boundary.  When we look at the internal
representation with Rmpfr,
we see a string of 9s.  the 44 at the end is just noise.  It looks
like the string of 99 is increased to 100
before rounding.
When we increase the precision from the default double precision
(precBits=53) to precBits=55, we
get the anticipated behavior.

> library(Rmpfr)
> round(.015, 2)
[1] 0.02
> getPrec(.015)
[1] 53
> mpfr(0.015, precBits=53)
1 'mpfr' number of precision  53   bits
[1] 0.015
> formatDec(mpfr(0.015, precBits=53))
[1] 0.014999999999999999
> round(0.014999999999999999, 2)
[1] 0.02
> round(0.014999999999999998, 2)
[1] 0.01
> > round(mpfr(0.015, precBits=54), 2)
1 'mpfr' number of precision  54   bits
[1] 0.02
> round(mpfr(0.015, precBits=55), 2)
1 'mpfr' number of precision  55   bits
[1] 0.01
> formatDec(mpfr(0.015, precBits=53))
[1] 0.014999999999999999
> formatDec(mpfr(0.015, precBits=54))
[1] 0.0149999999999999994
> formatDec(mpfr(0.015, precBits=55))
[1] 0.0149999999999999994
> roundMpfr(mpfr(0.014999999999999999, precBits=53), 53)
1 'mpfr' number of precision  53   bits
[1] 0.015
> roundMpfr(mpfr(0.014999999999999999, precBits=53), 54)
1 'mpfr' number of precision  54   bits
[1] 0.014999999999999999
> roundMpfr(mpfr(0.014999999999999999, precBits=53), 55)
1 'mpfr' number of precision  55   bits
[1] 0.014999999999999999
1
> formatHex(mpfr(0.015, precBits=53)*100)
[1] +0x1.8000000000000p+0
> formatHex(mpfr(0.015, precBits=54)*100)
[1] +0x1.80000000000000p+0
> formatHex(mpfr(0.015, precBits=55)*100)
[1] +0x1.7ffffffffffffcp+0
> formatHex(mpfr(0.015, precBits=53))
[1] +0x1.eb851eb851eb8p-7
> formatHex(mpfr(0.015, precBits=54))
[1] +0x1.eb851eb851eb80p-7
> formatHex(mpfr(0.015, precBits=55))
[1] +0x1.eb851eb851eb80p-7
> round(mpfr(0.015, precBits=53), 2)
1 'mpfr' number of precision  53   bits
[1] 0.02
> round(mpfr(0.015, precBits=54), 2)
1 'mpfr' number of precision  54   bits
[1] 0.02
> round(mpfr(0.015, precBits=55), 2)
1 'mpfr' number of precision  55   bits
[1] 0.01
>
>
On Wed, Nov 28, 2018 at 12:31 PM Philipp Upravitelev
<upravitelev using gmail.com> wrote:
>
> Dear colleagues,
> could you help me with the function base::round()? I can't understand how
> it works.
>
> For example, when I want to round 0.015 to the second digit, base::round()
> returns 0.02.
>
> But the real representation of the 0.015 is different:
> > sprintf('%.20f', 0.015)
> [1] "0.01499999999999999944"
> > 0.015 == 0.01499999999999999944
> [1] TRUE
> > round(0.015, 2)
> [1] 0.02
>
> Therefore, according to the arithmetic rules, rounded 0.014 to the second
> digit is 0.01. Also, the round() function in other programming languages
> (Python, Java) returns 0.01. It is a bit counterintuitive but
> mathematically correct.
>
> I'll be very pleased if you could help me to figure out why the
> base::round(0.015, 2) returns 0.02 and what is the purpose of this feature.
>
> Best regards,
> Philipp Upravitelev
>
>         [[alternative HTML version deleted]]
>
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