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

4 messages · Philipp Upravitelev, Rui Barradas, Duncan Murdoch +1 more

Philipp Upravitelev
# 
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:
[1] "0.01499999999999999944"
[1] TRUE
[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
Rui Barradas
# 
Hello,

Your assumption that you can sprintf with 20 digits of precision is 
wrong, you only have 16 decimal digits. And

sprintf('%.16f', 0.015)
#[1] "0.0150000000000000"

0.015 == 0.0150000000000000
#[1] TRUE

This rounds to the nearest even number, 0.02 (IEEE-754).

Hope this helps,

Rui Barradas

?s 13:49 de 28/11/2018, Philipp Upravitelev escreveu:
Duncan Murdoch
#
On 28/11/2018 8:49 AM, Philipp Upravitelev wrote:
This calculation is informative:

100*0.015 - 1.5

which gives 0 on my system.  So even though 0.015 isn't exactly 
representable, when you multiply by 100, you get the exactly correct 
result.  Then the apparent rule for round(x, 2) is:  multiply by 100, 
round to an integer, divide by 100.

Duncan Murdoch
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.
[1] 0.02
[1] 53
1 'mpfr' number of precision  53   bits
[1] 0.015
[1] 0.014999999999999999
[1] 0.02
[1] 0.01
1 'mpfr' number of precision  54   bits
[1] 0.02
1 'mpfr' number of precision  55   bits
[1] 0.01
[1] 0.014999999999999999
[1] 0.0149999999999999994
[1] 0.0149999999999999994
1 'mpfr' number of precision  53   bits
[1] 0.015
1 'mpfr' number of precision  54   bits
[1] 0.014999999999999999
1 'mpfr' number of precision  55   bits
[1] 0.014999999999999999
1
[1] +0x1.8000000000000p+0
[1] +0x1.80000000000000p+0
[1] +0x1.7ffffffffffffcp+0
[1] +0x1.eb851eb851eb8p-7
[1] +0x1.eb851eb851eb80p-7
[1] +0x1.eb851eb851eb80p-7
1 'mpfr' number of precision  53   bits
[1] 0.02
1 'mpfr' number of precision  54   bits
[1] 0.02
1 'mpfr' number of precision  55   bits
[1] 0.01
On Wed, Nov 28, 2018 at 12:31 PM Philipp Upravitelev
<upravitelev at gmail.com> wrote: