Python3如何以一定的精度向上舍入（向下）

Question

I need to round up a floating-point number. For example 4.00011 . The inbuilt function round() always rounds up when the number is > .5 and rounds down when <= 5. This is very good. When I want to round something up (down) I import math and use the function math.ceil() ( math.floor() ). The downside is that ceil() and floor() have no precision "settings". So as an R programmer I would basically just write my own function:

def my_round(x, precision = 0, which = "up"):   
    import math
    x = x * 10 ** precision
    if which == "up":
        x = math.ceil(x)
    elif which == "down":
        x = math.floor(x)
    x = x / (10 ** precision)
    return(x)

my_round(4.00018, 4, "up")

this prints 4.0002

my_round(4.00018, 4, "down")

this prints 4.0001

I can't find a question to this (why?). Is there any other module or function I've missed? Would be great to have a huge library with basic (altered) functions.

edit: I do not talk about integers.

Answer 1

Check out my answer from this SO post . You should be able to easily modify it to your needs by swapping floor for round .

Please let me know if that helps!

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EDIT I just felt it, so I wanted to propose a code based solution

import math

def round2precision(val, precision: int = 0, which: str = ''):
    assert precision >= 0
    val *= 10 ** precision
    round_callback = round
    if which.lower() == 'up':
        round_callback = math.ceil
    if which.lower() == 'down':
        round_callback = math.floor
    return '{1:.{0}f}'.format(precision, round_callback(val) / 10 ** precision)


quantity = 0.00725562
print(quantity)
print(round2precision(quantity, 6, 'up'))
print(round2precision(quantity, 6, 'down'))

which yields

0.00725562
0.007256
0.007255