getting Ceil() of Decimal in python?

问题:

Is there a way to get the ceil of a high precision Decimal in python?

>>> import decimal;
>>> decimal.Decimal(800000000000000000001)/100000000000000000000
Decimal('8.00000000000000000001')
>>> math.ceil(decimal.Decimal(800000000000000000001)/100000000000000000000)
8.0

math rounds the value and returns non precise value

回答1:

x = decimal.Decimal('8.00000000000000000000001')
with decimal.localcontext() as ctx:
    ctx.prec=100000000000000000
    ctx.rounding=decimal.ROUND_CEILING
    y = x.to_integral_exact()

回答2:

The most direct way to take the ceiling of a Decimal instance x is to use x.to_integral_exact(rounding=ROUND_CEILING). There's no need to mess with the context here. Note that this sets the Inexact and Rounded flags where appropriate; if you don't want the flags touched, use x.to_integral_value(rounding=ROUND_CEILING) instead. Example:

>>> from decimal import Decimal, ROUND_CEILING
>>> x = Decimal('-123.456')
>>> x.to_integral_exact(rounding=ROUND_CEILING)
Decimal('-123')

Unlike most of the Decimal methods, the to_integral_exact and to_integral_value methods aren't affected by the precision of the current context, so you don't have to worry about changing precision:

>>> from decimal import getcontext
>>> getcontext().prec = 2
>>> x.to_integral_exact(rounding=ROUND_CEILING)
Decimal('-123')

By the way, in Python 3.x, math.ceil works exactly as you want it to, except that it returns an int rather than a Decimal instance. That works because math.ceil is overloadable for custom types in Python 3. In Python 2, math.ceil simply converts the Decimal instance to a float first, potentially losing information in the process, so you can end up with incorrect results.

回答3:

You can do this using the precision and rounding mode option of the Context constructor.

ctx = decimal.Context(prec=1, rounding=decimal.ROUND_CEILING)
ctx.divide(decimal.Decimal(800000000000000000001), decimal.Decimal(100000000000000000000))

EDIT: You should consider changing the accepted answer.. Although the prec can be increased as needed, to_integral_exact is a simpler solution.

回答4:

>>> decimal.Context(rounding=decimal.ROUND_CEILING).quantize(
...   decimal.Decimal(800000000000000000001)/100000000000000000000, 0)
Decimal('9')

回答5:

def decimal_ceil(x):
    int_x = int(x)
    if x - int_x == 0:
        return int_x
    return int_x + 1

回答6:

Just use potency to make this. import math

def lo_ceil(num, potency=0): # Use 0 for multiples of 1, 1 for multiples of 10, 2 for 100 ...
      n = num / (10.0 ** potency)
      c = math.ceil(n)
      return c * (10.0 ** potency)

lo_ceil(8.0000001, 1) # return 10