Python3 How to round up(down) by a certain precision

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