问题
I would like to design a function f(x : float, up : bool) with these input/output:
# 2 decimals part rounded up (up = True)
f(142.452, True) = 142.46
f(142.449, True) = 142.45
# 2 decimals part rounded down (up = False)
f(142.452, False) = 142.45
f(142.449, False) = 142.44
Now, I know about Python's round built-in function but it will always round 142.449 up, which is not what I want.
Is there a way to do this in a nicer pythonic way than to do a bunch of float comparisons with epsilons (prone to errors)?
回答1:
Have you considered a mathematical approach using floor and ceil?
If you always want to round to 2 digits, then you could premultiply the number to be rounded by 100, then perform the rounding to the nearest integer and then divide again by 100.
from math import floor, ceil
def rounder(num, up=True):
digits = 2
mul = 10**digits
if up:
return ceil(num * mul)/mul
else:
return floor(num*mul)/mul
回答2:
You can also perform some mathematical logic if you do not want to use any explicit function as:
def f(num, up):
num = num * 100
if up and num != int(num): # if up and "float' value != 'int' value
num += 1
return int(num) / (100.0)
Here, the idea is if up is True and int value of number is not equal to float value then increase the number by 1. Else it will be same as the original number
回答3:
math.ceil() rounds up, and math.floor() rounds down. So, the following is an example of how to use it:
import math
def f(x, b):
if b:
return (math.ceil(100*x) / 100)
else:
return (math.floor(100*x) / 100)
This function should do exactly what you want.
来源:https://stackoverflow.com/questions/40613590/round-up-down-float-to-2-decimals