我刚刚重读了Python 3.0的新特性,它说:

round()函数的舍入策略和返回类型已更改。 精确的中间情况现在四舍五入到最接近的偶数结果代替 远离0的。(例如,round(2.5)现在返回2而不是 3)。

和 round的文档:

对于支持round()的内置类型,值舍入为 10的- n次方最接近的倍数;如果两个倍数是 同样接近,四舍五入是对偶数的选择

因此,在v2.7.3中:

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

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

正如我所预料的。然而,现在在v3.2.3下:

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

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

这似乎有违直觉,与我的理解相悖 四舍五入(肯定会绊倒人)。英语不是我的母语,但是 直到我读到这篇文章,我以为我知道四舍五入的意思:-/我确定 在v3被引入的时候,肯定有一些讨论 但在我的寻找中,我找不到一个好的理由。

有人知道为什么这个变成了这个吗? 有没有其他主流编程语言(例如,C, c++, Java, Perl, ..)做这种类型的舍入(对我来说不一致)?

我错过了什么?

更新:@Li-aungYip关于“银行家的舍入”的评论给了我正确的搜索词/关键字来搜索,我发现了这个SO问题:为什么。net使用银行家的舍入作为默认值?,所以我会仔细阅读。


当前回答

Python 3。X将.5值舍入为偶数的相邻值

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

但是,如果需要,可以将十进制舍入“向后”更改为始终向上舍入。5:

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

其他回答

# 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)

Python 3的方法(称为“四舍五入”或“银行家四舍五入”)目前被认为是标准的四舍五入方法,尽管一些语言实现还没有在总线上实现。

简单的“总是四舍五入0.5”技术会导致对较大数字的轻微偏向。对于大量的计算,这是非常重要的。Python 3.0方法消除了这个问题。

常用的四舍五入方法不止一种。IEEE 754是浮点数学的国际标准,定义了五种不同的舍入方法(Python 3.0使用的是默认方法)。还有其他的。

这种行为并没有像它应有的那样广为人知。如果我没记错的话,AppleScript是这种舍入方法的早期采用者。AppleScript中的round命令提供了几个选项,但在IEEE 754中,round-向偶数是默认的。显然,实现round命令的工程师受够了“让它像我在学校学到的那样工作”的所有请求,所以他实现了:学校教的round 2.5舍入是一个有效的AppleScript命令。: -)

Python 3。X将.5值舍入为偶数的相邻值

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

但是,如果需要,可以将十进制舍入“向后”更改为始终向上舍入。5:

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

我建议自定义函数,它将工作于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)

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)

希望这能有所帮助,

Narnie