Python - round up to the nearest ten

Question

If I get the number 46 and I want to round up to the nearest ten. How do can I do this in python?

46 goes to 50.

Nearest *ten* (`10`) or nearest *tenth* (`0.1`)? What have you tried, and what exactly is the problem with it? — jonrsharpe, Oct 19 '14 at 19:49
What type of numbers? `int`s? `float`s? `Decimal` instances? — Mark Dickinson, Oct 20 '14 at 17:44

score 184 · Answer 1 · answered Oct 19 '14 at 19:51

184

round does take negative ndigits parameter!

>>> round(46,-1)
50

may solve your case.

answered Oct 19 '14 at 19:51

ch3ka

13

This doesn't round *up* though. It also returns a `float`, which may or may not be desirable (but I think is worth mentioning). – NPE Oct 19 '14 at 19:53
9

true, it rounds to the nearest... on py3, it returns an int though. – ch3ka Oct 19 '14 at 19:53
2

You can easily enough put it inside of int() if you don't want a float. Thanks. This solved my case. – EL_DON Aug 24 '16 at 05:19
7

It's dirty but you could always do `round(num + 5.1, -1)` – Jan 08 '18 at 22:07
2

bests answer right here – Brainmaniac Jul 06 '18 at 09:37

Parker · Accepted Answer · 2023-05-17T14:25:52.680

91

You can use math.ceil() to round up, and then multiply by 10

Python 2

import math

def roundup(x):
    return int(math.ceil(x / 10.0)) * 10

Python 3 (the only difference is you no longer need to cast the result as an int)

import math

def roundup(x):
    return math.ceil(x / 10.0) * 10

To use just do

>>roundup(45)
50

edited May 17 '23 at 14:25

answered Oct 19 '14 at 19:58

Parker

That does not work always: If `x = 1000000000000000010` then `print(int(math.ceil(x / 10.0)) * 10)` outputs `1000000000000000000`. – Jaakko Seppälä Nov 13 '17 at 18:55
3

I think that's because you're going past the python int limit – Parker Dec 11 '17 at 01:09
As far as I know, there is no longer a limit on the maximum size of an integer in Python, see here: https://www.daniweb.com/programming/software-development/threads/71008/comparing-python-and-c-part-1-integer-variables – Martin Hepp Nov 07 '19 at 19:45
It is more likely because you hit the `eps` and finite precision take over after that. – sotmot Dec 18 '21 at 15:29
1

The expression `math.ceil(x / 10) * 10` works just fine. No need for those conversions in recent versions of Python. – Jeyekomon Feb 08 '22 at 12:52

score 26 · Answer 3 · answered Oct 19 '14 at 19:50

26

Here is one way to do it:

>>> n = 46
>>> (n + 9) // 10 * 10
50

answered Oct 19 '14 at 19:50

NPE

5

Nice trick. Boils down to adding the numbers base minus 1. – Mads Y Dec 05 '15 at 16:45
2

this would fail for numbers like 0.5 or 30.5 and so on (i.e. k*10+a where k is an integer and 0 < a < 1 ) – epeleg Oct 03 '17 at 09:52
3

yes the answer assumes the input in an integer only - not clear from the question that this was a condition, although the example implies this is the case – ClimateUnboxed Oct 09 '18 at 01:58

score 16 · Answer 4 · edited Nov 06 '15 at 13:08

16

This will round down correctly as well:

>>> n = 46
>>> rem = n % 10
>>> if rem < 5:
...     n = int(n / 10) * 10
... else:
...     n = int((n + 10) / 10) * 10
...
>>> 50

edited Nov 06 '15 at 13:08

Community

answered Oct 19 '14 at 20:02

primussucks

4

The requirement is specifically to round *up*. – NPE Oct 19 '14 at 20:12
5

It helps to have the round down too. Thanks jvasilakes – Nav Apr 22 '19 at 04:05

4 Answers4