Python 3中的舍入行为。

小数点后15位加1。 精度可达15位。

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

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

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)

你可以使用Decimal模块控制你在Py3000中得到的舍入:

>>> 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')

试试下面的代码:

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))

结果将是:

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

Ooutput你可以在这里检查: https://i.stack.imgur.com/QQUkS.png

Python 3中最简单的舍入方法。学校教的X使用辅助变量:

n = 0.1 
round(2.5 + n)

这些将是2.0到3.0系列的结果(0.1步):

>>> round(2 + n)
>>> 2

>>> round(2.1 + n)
>>> 2

>>> round(2.2 + n)
>>> 2

>>> round(2.3 + n)
>>> 2

>>> round(2.4 + n)
>>> 2

>>> round(2.5 + n)
>>> 3

>>> round(2.6 + n)
>>> 3

>>> round(2.7 + n)
>>> 3

>>> round(2.8 + n)
>>> 3

>>> round(2.9 + n)
>>> 3

>>> round(3 + n)
>>> 3