Rounding a number to nearest even

[bdsatish]

The built-in function round( ) will always "round up", that is 1.5 is
rounded to 2.0 and 2.5 is rounded to 3.0.

If I want to round to the nearest even, that is

my_round(1.5) = 2 # As expected
my_round(2.5) = 2 # Not 3, which is an odd num

I'm interested in rounding numbers of the form "x.5" depending upon
whether x is odd or even. Any idea about how to implement it ?

[colas.francis@gmail.com]

Re: Rounding a number to nearest even

On 11 avr, 12:14, bdsatish <bdsat...@gmail .comwrote:

The built-in function round( ) will always "round up", that is 1.5 is
rounded to 2.0 and 2.5 is rounded to 3.0.
>
If I want to round to the nearest even, that is
>
my_round(1.5) = 2 # As expected
my_round(2.5) = 2 # Not 3, which is an odd num
>
I'm interested in rounding numbers of the form "x.5" depending upon
whether x is odd or even. Any idea about how to implement it ?

When you say "round to the nearest even", you mean new_round(3) <3?

Is so, you can try:

In [37]: def new_round(x):
....: return round(x/2.)*2
....:

In [38]: new_round(1.5)
Out[38]: 2.0

In [39]: new_round(2.5)
Out[39]: 2.0

In [40]: new_round(3)
Out[40]: 4.0

[bdsatish]

Re: Rounding a number to nearest even

On Apr 11, 3:27 pm, colas.fran...@g mail.com wrote:

On 11 avr, 12:14, bdsatish <bdsat...@gmail .comwrote:
>

The built-in function round( ) will always "round up", that is 1.5 is
rounded to 2.0 and 2.5 is rounded to 3.0.

>

If I want to round to the nearest even, that is

>

my_round(1.5) = 2 # As expected
my_round(2.5) = 2 # Not 3, which is an odd num

>

I'm interested in rounding numbers of the form "x.5" depending upon
whether x is odd or even. Any idea about how to implement it ?

>
When you say "round to the nearest even", you mean new_round(3) <3?

No. not at all. The clause "nearest even" comes into picture only when
a number is of form "x.5" or else it's same as builtin round( ).
new_round(3.0) must be 3.0 itself. Here is the mathematical definition
of what I want:

If 'n' is an integer,

new_round(n+0.5 ) = n if n/2 is integer
new_round(n+0.5 ) = (n+1) if (n+1)/2 is integer

In all other cases, new_round() behave similarly as round( ). Here are
the results I expect:

new_round(3.2) = 3
new_round(3.6) = 4
new_round(3.5) = 4
new_round(2.5) = 2
new_round(-0.5) = 0.0
new_round(-1.5) = -2.0
new_round(-1.3) = -1.0
new_round(-1.8) = -2
new_round(-2.5) = -2.0

The built-in function doesnt meet my needs for round(-2.5) or
round(2.5)

[cokofreedom@gmail.com]

Re: Rounding a number to nearest even

couldn't you just do.

#untested
new_round(n):
answer = round(n)
# is answer now odd
if answer % 2:
return answer - 1
else:
return answer

[cokofreedom@gmail.com]

Re: Rounding a number to nearest even

On Apr 11, 1:19 pm, cokofree...@gma il.com wrote:

couldn't you just do.
>
#untested
new_round(n):
answer = round(n)
# is answer now odd
if answer % 2:
return answer - 1
else:
return answer

Whoops, this also affects odd numbers...

Will try and find a GOOD solution later...

Strange request though, why do you need it that way, because 2.5 is
CLOSER to 3 than to 2...

[bdsatish]

Re: Rounding a number to nearest even

On Apr 11, 4:19 pm, cokofree...@gma il.com wrote:

couldn't you just do.
>
#untested
new_round(n):
answer = round(n)
# is answer now odd
if answer % 2:
return answer - 1
else:
return answer

It fails for negative numbers: For -2.5 it gives -4.0 as answer
whereas I expect -2.0

[Scott David Daniels]

Re: Rounding a number to nearest even

bdsatish wrote:

The built-in function round( ) will always "round up", that is 1.5 is
rounded to 2.0 and 2.5 is rounded to 3.0.
>
If I want to round to the nearest even, that is
>
my_round(1.5) = 2 # As expected
my_round(2.5) = 2 # Not 3, which is an odd num
>
I'm interested in rounding numbers of the form "x.5" depending upon
whether x is odd or even. Any idea about how to implement it ?

def rounded(v):
rounded = round(v)
if divmod(v, 1)[1] == .5 and divmod(rounded, 2)[1] == 1:
if v 0:
return rounded - 1
return rounded + 1
return rounded

last = None
for n in range(-29, 28):
x = n * .25
r = xr(x)
if r != last:
last = r
print
print '%s->%s' % (x, xr(x)),

-7.25->-7.0 -7.0->-7.0 -6.75->-7.0
-6.5->-6.0 -6.25->-6.0 -6.0->-6.0 -5.75->-6.0 -5.5->-6.0
-5.25->-5.0 -5.0->-5.0 -4.75->-5.0
-4.5->-4.0 -4.25->-4.0 -4.0->-4.0 -3.75->-4.0 -3.5->-4.0
-3.25->-3.0 -3.0->-3.0 -2.75->-3.0
-2.5->-2.0 -2.25->-2.0 -2.0->-2.0 -1.75->-2.0 -1.5->-2.0
-1.25->-1.0 -1.0->-1.0 -0.75->-1.0
-0.5->0.0 -0.25->-0.0 0.0->0.0 0.25->0.0 0.5->0.0
0.75->1.0 1.0->1.0 1.25->1.0
1.5->2.0 1.75->2.0 2.0->2.0 2.25->2.0 2.5->2.0
2.75->3.0 3.0->3.0 3.25->3.0
3.5->4.0 3.75->4.0 4.0->4.0 4.25->4.0 4.5->4.0
4.75->5.0 5.0->5.0 5.25->5.0
5.5->6.0 5.75->6.0 6.0->6.0 6.25->6.0 6.5->6.0
6.75->7.0

[bdsatish]

Re: Rounding a number to nearest even

On Apr 11, 4:24 pm, cokofree...@gma il.com wrote:

On Apr 11, 1:19 pm, cokofree...@gma il.com wrote:
>

couldn't you just do.

>

#untested
new_round(n):
answer = round(n)
# is answer now odd
if answer % 2:
return answer - 1
else:
return answer

>
Whoops, this also affects odd numbers...
>
Will try and find a GOOD solution later...
>
Strange request though, why do you need it that way, because 2.5 is
CLOSER to 3 than to 2...

It also fails for negative numbers. For -2.5 as input, I get -4.0
whereas I expect -2.0

This is a lengthy solution I came-up with:

def round_even(x):
temp = round(abs(x))
if (abs(x) - 0.5)%2.0 == 0.0: temp=temp-1
return signum(x)*temp

def signum(x):
if x>0: return 1
if x<0: return -1
return 0

But i guess there are better ways. I need it 'cos I'm translating some
code from Mathematica to Python. And Math..ica's Round[ ] behaves this
way (as I requested)

[bdsatish]

Re: Rounding a number to nearest even

On Apr 11, 4:37 pm, Scott David Daniels <Scott.Dani...@ Acm.Orgwrote:

bdsatish wrote:

The built-in function round( ) will always "round up", that is 1.5 is

def rounded(v):
rounded = round(v)
if divmod(v, 1)[1] == .5 and divmod(rounded, 2)[1] == 1:
if v 0:
return rounded - 1
return rounded + 1
return rounded
>
last = None
for n in range(-29, 28):
x = n * .25
r = xr(x)
if r != last:
last = r
print
print '%s->%s' % (x, xr(x)),
>

Hi Scott,
This is what I was looking for.. I forgot about divmod( ) thanks for
reminding.

[Ivan Illarionov]

Re: Rounding a number to nearest even

On Apr 11, 2:14 pm, bdsatish <bdsat...@gmail .comwrote:

The built-in function round( ) will always "round up", that is 1.5 is
rounded to 2.0 and 2.5 is rounded to 3.0.
>
If I want to round to the nearest even, that is
>
my_round(1.5) = 2 # As expected
my_round(2.5) = 2 # Not 3, which is an odd num
>
I'm interested in rounding numbers of the form "x.5" depending upon
whether x is odd or even. Any idea about how to implement it ?

def even_round(x):
if x % 1 == .5 and not (int(x) % 2):
return float(int(x))
else:
return round(x)

nums = [ 3.2, 3.6, 3.5, 2.5, -.5, -1.5, -1.3, -1.8, -2.5 ]
for num in nums:
print num, '->', even_round(num)

3.2 -3.0
3.6 -4.0
3.5 -4.0
2.5 -2.0
-0.5 -0.0
-1.5 --2.0
-1.3 --1.0
-1.8 --2.0
-2.5 --2.0

[Graham Breed]

Re: Rounding a number to nearest even

On Apr 11, 6:14 pm, bdsatish <bdsat...@gmail .comwrote:

The built-in function round( ) will always "round up", that is 1.5 is
rounded to 2.0 and 2.5 is rounded to 3.0.
>
If I want to round to the nearest even, that is
>
my_round(1.5) = 2 # As expected
my_round(2.5) = 2 # Not 3, which is an odd num

If you care about such details, you may be better off using decimals
instead of floats.

I'm interested in rounding numbers of the form "x.5" depending upon
whether x is odd or even. Any idea about how to implement it ?

import decimal
decimal.Decimal ("1.5").to_inte gral(
rounding=decima l.ROUND_HALF_EV EN)
decimal.Decimal ("2.5").to_inte gral(
rounding=decima l.ROUND_HALF_EV EN)

ROUND_HALF_EVEN is the default, but maybe that can be changed, so
explicit is safest.

If you really insist,

import decimal
def my_round(f):
d = decimal.Decimal (str(f))
rounded = d.to_integral(r ounding=decimal .ROUND_HALF_EVE N)
return int(rounded)


Graham

[Robert Kern]

Re: Rounding a number to nearest even

cokofreedom@gma il.com wrote:

Strange request though, why do you need it that way, because 2.5 is
CLOSER to 3 than to 2...

Uhhh, no it isn't. (3 - 2.5) == (2.5 - 2)

--
Robert Kern

"I have come to believe that the whole world is an enigma, a harmless enigma
that is made terrible by our own mad attempt to interpret it as though it had
an underlying truth."
-- Umberto Eco

[Mark Dickinson]

Re: Rounding a number to nearest even

On Apr 11, 10:29 am, hdante <hda...@gmail.c omwrote:

Strangely, a "faster" version is:
>
def fast_round(x):
if x % 1 != 0.5: return round(x)
return 2.0*round(x/2.0)

You should be a little bit careful with the test
x%1 == 0.5 if x might be negative:

>>x = -0.5 + 2**-54
>>x # not an exact half...

-0.4999999999999 9994

>>x % 1 # ... and yet x%1 == 0.5

0.5

But for x positive, it should be safe. And for this
particular application, it turns out that it doesn't
matter: it gives the right result for this input anyway.

Mark

[Mark Dickinson]

Re: Rounding a number to nearest even

On Apr 11, 2:33 pm, Lie <Lie.1...@gmail .comwrote:

In this table, we consider that a number is rounded down when the
number is equal to truncated value (the number without fractional
part), while round up is equal to truncated value + 1 or truncated
value -1 if value is negative (Actually this is not round-half-up
algorithm, it's a round-half-away-from-zero algorithm, but lets just
consider that to be the same for now). In this revised table, you get
10 round ups and 10 round down (i.e. Average Rounding Error == 0),
while by rounding to nearest even you get 9 round up and 11 round down
(i.e. Average Rounding Error != 0).

No: if you interpret average to mean 'mean' (add up the (signed)
rounding errors from the 20 cases you list and divide by 20) you'll
find that the average rounding error is indeed 0 for round-half-to-
even,
and 0.05 for round-half-away-from-0, just as Mikael said.

Another mistake, in an unquantized value the probability of getting
exactly 0.5 (or any other number specified) is not 0 but an
infinitesimal (i.e. lim(x) where x -0 (BUT NOT ZERO))

I'm not sure you'll get many takers for this point of view. If X is
a random variable uniformly distributed on the interval [0, 1) then
the probability that X == 0.5 is indeed exactly 0, using conventional
definitions. (And if you're not using conventional definitions, you
should say so....)

Mark

[Arnaud Delobelle]

Re: Rounding a number to nearest even

On Apr 11, 8:27 pm, Mark Dickinson <dicki...@gmail .comwrote:

On Apr 11, 2:33 pm, Lie <Lie.1...@gmail .comwrote:

[...]

Another mistake, in an unquantized value the probability of getting
exactly 0.5 (or any other number specified) is not 0 but an
infinitesimal (i.e. lim(x) where x -0 (BUT NOT ZERO))

>
I'm not sure you'll get many takers for this point of view.  If X is
a random variable uniformly distributed on the interval [0, 1) then
the probability that X == 0.5 is indeed exactly 0, using conventional
definitions.  (And if you're not using conventional definitions, you
should say so....)

And I would like to know what unconventional - but mathematically
meaningful - definitions lead to

lim x != 0
x -0

--
Arnaud