半圆形实施

Question

Is there a way in Python to round the number up only if the fractional part is larger than a half? round and math.ceil will do this:

round(X.5) => X+1
math.ceil(X.5) => X+1

however, I want this:

x.500...000001 => x+1
x.500...000000 => x

How to do that?

I have 2.7.13 python x64

Answer 1

math.ceil(x-0.5)

maps:

12.50 -> ceil(12.00) -> 12
12.49 -> ceil(11.99) -> 12
12.51 -> ceil(12.01) -> 13

Answer 2

You can do something like this:

round(X - 0.00000000001)

You will need to decide how many decimal places are appropriate, bearing in mind that floats are not exact, so for example 10000000000000000000.5 can't be represented.

Answer 3

I found the answer as suggested by jonrsharpe to use context This is they way I used it:

from decimal import *
x =5.500000000000001   # it is the number wanted to round, however only takes 15 number after decimal for me
ctx= Context(prec=1, rounding=ROUND_HALF_DOWN)
y = ctx.create_decimal(x)  

However, I do not know if it is going to work with everybody. probably there is a better way to put it.

Answer 4

import math
while True:
    num=input("num: ")
    num=float(num)
    new_n=round(num,0)
    if (float(new_n)-0.5)==num:
        print(num-0.5)
    else:
        print(new_n)

output:

num: 4.5
4.0
num: 5.5
5.0
num: 6.6
7.0
num: 5.5001
6.0

well I had to do some thinking but this should do what you want