Python 3.x rounding behavior

Question

I was just re-reading What’s New In Python 3.0 and it states:

The round() function rounding strategy and return type have changed. Exact halfway cases are now rounded to the nearest even result instead of away from zero. (For example, round(2.5) now returns 2 rather than 3.)

and the documentation for round:

For the built-in types supporting round(), values are rounded to the closest multiple of 10 to the power minus n; if two multiples are equally close, rounding is done toward the even choice

So, under v2.7.3:

In [85]: round(2.5)
Out[85]: 3.0

In [86]: round(3.5)
Out[86]: 4.0

as I'd have expected. However, now under v3.2.3:

In [32]: round(2.5)
Out[32]: 2

In [33]: round(3.5)
Out[33]: 4

This seems counter-intuitive and contrary to what I understand about rounding (and bound to trip up people). English isn't my native language but until I read this I thought I knew what rounding meant :-/ I am sure at the time v3 was introduced there must have been some discussion of this, but I was unable to find a good reason in my search.

1. Does anyone have insight into why this was changed to this?
2. Are there any other mainstream programming languages (e.g., C, C++, Java, Perl, ..) that do this sort of (to me inconsistent) rounding?

What am I missing here?

UPDATE: @Li-aungYip's comment re "Banker's rounding" gave me the right search term/keywords to search for and I found this SO question: Why does .NET use banker's rounding as default?, so I will be reading that carefully.

Answer 1

Python 3's way (called "round half to even" or "banker's rounding") is considered the standard rounding method these days, though some language implementations aren't on the bus yet.

The simple "always round 0.5 up" technique results in a slight bias toward the higher number. With large numbers of calculations, this can be significant. The Python 3.0 approach eliminates this issue.

There is more than one method of rounding in common use. IEEE 754, the international standard for floating-point math, defines five different rounding methods (the one used by Python 3.0 is the default). And there are others.

This behavior is not as widely known as it ought to be. AppleScript was, if I remember correctly, an early adopter of this rounding method. The round command in AppleScript offers several options, but round-toward-even is the default as it is in IEEE 754. Apparently the engineer who implemented the round command got so fed up with all the requests to "make it work like I learned in school" that he implemented just that: round 2.5 rounding as taught in school is a valid AppleScript command. :-)

Answer 2

You can control the rounding you get in Py3000 using the Decimal module:

>>> decimal.Decimal('3.5').quantize(decimal.Decimal('1'), 
    rounding=decimal.ROUND_HALF_UP)
>>> Decimal('4')

>>> decimal.Decimal('2.5').quantize(decimal.Decimal('1'),    
    rounding=decimal.ROUND_HALF_EVEN)
>>> Decimal('2')

>>> decimal.Decimal('3.5').quantize(decimal.Decimal('1'), 
    rounding=decimal.ROUND_HALF_DOWN)
>>> Decimal('3')

Answer 3

Just to add here an important note from documentation:

https://docs.python.org/dev/library/functions.html#round

Note

The behavior of round() for floats can be surprising: for example, round(2.675, 2) gives 2.67 instead of the expected 2.68. This is not a bug: it’s a result of the fact that most decimal fractions can’t be represented exactly as a float. See Floating Point Arithmetic: Issues and Limitations for more information.

So don't be surprised to get following results in Python 3.2:

>>> round(0.25,1), round(0.35,1), round(0.45,1), round(0.55,1)
(0.2, 0.3, 0.5, 0.6)

>>> round(0.025,2), round(0.035,2), round(0.045,2), round(0.055,2)
(0.03, 0.04, 0.04, 0.06)

Answer 4

Python 3.x rounds .5 values to a neighbour which is even

assert round(0.5) == 0
assert round(1.5) == 2
assert round(2.5) == 2

import decimal

assert decimal.Decimal('0.5').to_integral_value() == 0
assert decimal.Decimal('1.5').to_integral_value() == 2
assert decimal.Decimal('2.5').to_integral_value() == 2

however, one can change decimal rounding "back" to always round .5 up, if needed :

decimal.getcontext().rounding = decimal.ROUND_HALF_UP

assert decimal.Decimal('0.5').to_integral_value() == 1
assert decimal.Decimal('1.5').to_integral_value() == 2
assert decimal.Decimal('2.5').to_integral_value() == 3

i = int(decimal.Decimal('2.5').to_integral_value()) # to get an int
assert i == 3
assert type(i) is int

Answer 5

I recently had problems with this, too. Hence, I have developed a python 3 module that has 2 functions trueround() and trueround_precision() that address this and give the same rounding behaviour were are used to from primary school (not banker's rounding). Here is the module. Just save the code and copy it in or import it. Note: the trueround_precision module can change the rounding behaviour depending on needs according to the ROUND_CEILING, ROUND_DOWN, ROUND_FLOOR, ROUND_HALF_DOWN, ROUND_HALF_EVEN, ROUND_HALF_UP, ROUND_UP, and ROUND_05UP flags in the decimal module (see that modules documentation for more info). For the functions below, see the docstrings or use help(trueround) and help(trueround_precision) if copied into an interpreter for further documentation.

#! /usr/bin/env python3
# -*- coding: utf-8 -*-

def trueround(number, places=0):
    '''
    trueround(number, places)
    
    example:
        
        >>> trueround(2.55, 1) == 2.6
        True

    uses standard functions with no import to give "normal" behavior to 
    rounding so that trueround(2.5) == 3, trueround(3.5) == 4, 
    trueround(4.5) == 5, etc. Use with caution, however. This still has 
    the same problem with floating point math. The return object will 
    be type int if places=0 or a float if places=>1.
    
    number is the floating point number needed rounding
    
    places is the number of decimal places to round to with '0' as the
        default which will actually return our interger. Otherwise, a
        floating point will be returned to the given decimal place.
    
    Note:   Use trueround_precision() if true precision with
            floats is needed

    GPL 2.0
    copywrite by Narnie Harshoe <signupnarnie@gmail.com>
    '''
    place = 10**(places)
    rounded = (int(number*place + 0.5if number>=0 else -0.5))/place
    if rounded == int(rounded):
        rounded = int(rounded)
    return rounded

def trueround_precision(number, places=0, rounding=None):
    '''
    trueround_precision(number, places, rounding=ROUND_HALF_UP)
    
    Uses true precision for floating numbers using the 'decimal' module in
    python and assumes the module has already been imported before calling
    this function. The return object is of type Decimal.

    All rounding options are available from the decimal module including 
    ROUND_CEILING, ROUND_DOWN, ROUND_FLOOR, ROUND_HALF_DOWN, ROUND_HALF_EVEN, 
    ROUND_HALF_UP, ROUND_UP, and ROUND_05UP.

    examples:
        
        >>> trueround(2.5, 0) == Decimal('3')
        True
        >>> trueround(2.5, 0, ROUND_DOWN) == Decimal('2')
        True

    number is a floating point number or a string type containing a number on 
        on which to be acted.

    places is the number of decimal places to round to with '0' as the default.

    Note:   if type float is passed as the first argument to the function, it
            will first be converted to a str type for correct rounding.

    GPL 2.0
    copywrite by Narnie Harshoe <signupnarnie@gmail.com>
    '''
    from decimal import Decimal as dec
    from decimal import ROUND_HALF_UP
    from decimal import ROUND_CEILING
    from decimal import ROUND_DOWN
    from decimal import ROUND_FLOOR
    from decimal import ROUND_HALF_DOWN
    from decimal import ROUND_HALF_EVEN
    from decimal import ROUND_UP
    from decimal import ROUND_05UP

    if type(number) == type(float()):
        number = str(number)
    if rounding == None:
        rounding = ROUND_HALF_UP
    place = '1.'
    for i in range(places):
        place = ''.join([place, '0'])
    return dec(number).quantize(dec(place), rounding=rounding)

Answer 6

Python 2 rounding behaviour in python 3.

Adding 1 at the 15th decimal places. Accuracy upto 15 digits.

round2=lambda x,y=None: round(x+1e-15,y)

Not right for 175.57. For that it should be added in the 13th decimal place as the number is grown. Switching to Decimal is better than reinventing the same wheel.

from decimal import Decimal, ROUND_HALF_UP
def round2(x, y=2):
    prec = Decimal(10) ** -y
    return float(Decimal(str(round(x,3))).quantize(prec, rounding=ROUND_HALF_UP))

Not used y

Answer 7

Some cases:

in: Decimal(75.29 / 2).quantize(Decimal('0.01'), rounding=ROUND_HALF_UP)
in: round(75.29 / 2, 2)
out: 37.65 GOOD

in: Decimal(85.55 / 2).quantize(Decimal('0.01'), rounding=ROUND_HALF_UP)
in: round(85.55 / 2, 2)
out: 42.77 BAD

For fix:

in: round(75.29 / 2 + 0.00001, 2)
out: 37.65 GOOD
in: round(85.55 / 2 + 0.00001, 2)
out: 42.78 GOOD

If you want more decimals, for example 4, you should add (+ 0.0000001).

Work for me.

Answer 8

Sample Reproduction:

['{} => {}'.format(x+0.5, round(x+0.5)) for x in range(10)]

['0.5 => 0', '1.5 => 2', '2.5 => 2', '3.5 => 4', '4.5 => 4', '5.5 => 6', '6.5 => 6', '7.5 => 8', '8.5 => 8', '9.5 => 10']

API: https://docs.python.org/3/library/functions.html#round

States:

Return number rounded to ndigits precision after the decimal point. If ndigits is omitted or is None, it returns the nearest integer to its input.

For the built-in types supporting round(), values are rounded to the closest multiple of 10 to the power minus ndigits; if two multiples are equally close, rounding is done toward the even choice (so, for example, both round(0.5) and round(-0.5) are 0, and round(1.5) is 2). Any integer value is valid for ndigits (positive, zero, or negative). The return value is an integer if ndigits is omitted or None. Otherwise the return value has the same type as number.

For a general Python object number, round delegates to number.round.

Note The behavior of round() for floats can be surprising: for example, round(2.675, 2) gives 2.67 instead of the expected 2.68. This is not a bug: it’s a result of the fact that most decimal fractions can’t be represented exactly as a float. See Floating Point Arithmetic: Issues and Limitations for more information.

Given this insight you can use some math to resolve it

import math
def my_round(i):
  f = math.floor(i)
  return f if i - f < 0.5 else f+1

now you can run the same test with my_round instead of round.

['{} => {}'.format(x + 0.5, my_round(x+0.5)) for x in range(10)]
['0.5 => 1', '1.5 => 2', '2.5 => 3', '3.5 => 4', '4.5 => 5', '5.5 => 6', '6.5 => 7', '7.5 => 8', '8.5 => 9', '9.5 => 10']

Answer 9

I propose custom function which would work for a DataFrame:

def dfCustomRound(df, dec):
    d = 1 / 10 ** dec
    df = round(df, dec + 2)
    return (((df % (1 * d)) == 0.5 * d).astype(int) * 0.1 * d * np.sign(df) + df).round(dec)

Answer 10

# round module within numpy when decimal is X.5 will give desired (X+1)

import numpy as np
example_of_some_variable = 3.5
rounded_result_of_variable = np.round(example_of_some_variable,0)
print (rounded_result_of_variable)

Answer 11

Try this code:

def roundup(input):   
   demo = input  if str(input)[-1] != "5" else str(input).replace("5","6")
   place = len(demo.split(".")[1])-1
   return(round(float(demo),place))

The result will be:

>>> x = roundup(2.5)
>>> x
3.0  
>>> x = roundup(2.05)
>>> x
2.1 
>>> x = roundup(2.005)
>>> x
2.01 

Ooutput you can check here: https://i.stack.imgur.com/QQUkS.png

Answer 12

You can control the rounding you using the math.ceil module:

import math
print(math.ceil(2.5))
> 3