Round last digit integer to nearest 9

Question

Set-up

I have a set of prices with decimals I want to end on the nearest 9 , where the cut-off point is 4 .

Eg 734 should be 729 , 734.1 should be 739 and 733.9 should be 729 .

Prices can be single, double, triple, and quadruple digits with decimals.

---

Try

if int(str(int(757.2))[-1]) > 4:
    p = int(str(int(757.2))[:-1] + '9')

Now, this returns input price 757.2 as 759 , as desired. However, if the input price would be ending on a 3 and/or wouldn't be a triple-digit I wouldn't know how to account for it.

Also, this approach seems rather cumbersome.

Is there someone who knows how to fix this neatly?

Answer 1

Just do the typical round to 10 and shift by 1 before and after:

def round9(n):
    return round((n+1) / 10) * 10 - 1

round9(737.2)
# 739
round9(733.2)
# 729
round9(734.0)
# 739
round9(733.9)
# 729

Answer 2

IIUC, try:

import math
def nearest(n):
    if n%10 > 4:
        return math.ceil(n/10)*10 - 1
    return math.floor(n/10)*10 - 1

>>> nearest(734)
729

>>> nearest(734.1)
739

>>> nearest(733.9)
729

Answer 3

You could get the integer division by 10 with a condition on the remainder, and multiply by ten to get the nearest 10, then subtract 1:

def nearest9(n):
    d,r = divmod(n,10)
    return int(d+(r>4))*10-1

>>> nearest9(734.1)
739

>>> nearest9(734)
729