How to round decimal digits always up

Question

So i ran into trouble rounding floats up, this is my code:

foo = float(0.21)
bar = float(0.871929)

foobar = foo * bar
Rfoobar = round(foobar,2)

This gives me:

foobar = 0.1831
Rfoobar = 0.18

But i want Rfoobar to be 0.19, how do i accomplish that it always rounds up the digits when there is a remainder?

I read about math.ceiling but in my case that doesn't seem to do the trick.

any help is greatly appreciated.

@Tom Zych You should put that in an answer, because it's totally what I would have done if not for your comment. ;-) — John Sensebe, Jan 24 '16 at 19:46
You've read [_how to round off a floating number in python_](http://stackoverflow.com/questions/4518641/how-to-round-off-a-floating-number-in-python)? — martineau, Jan 24 '16 at 19:50
@TomZych How come i didn't think of that duh, i need more coffee! Thanks for the help, that works! — Lyux, Jan 24 '16 at 19:52

score 2 · Accepted Answer · answered Jan 24 '16 at 19:50

2

Just move the decimal point before and after calling ceil:

from math import ceil

Rfoobar = ceil(foobar * 100) / 100

If the number of decimals varies, you can do something like:

Rfoobar = ceil(foobar * 10 ** digits) / 10 ** digits

answered Jan 24 '16 at 19:50

Tom Zych

score 2 · Answer 2 · answered Jan 24 '16 at 20:27

2

You could also add 0.005 before rounding:

>>> for foobar in 0.1831, 0.1801, 0.1800:
        print(foobar, '->', round(foobar + 0.005, 2))

0.1831 -> 0.19
0.1801 -> 0.19
0.18 -> 0.18

answered Jan 24 '16 at 20:27

Stefan Pochmann

2 Answers2