Round integers to the nearest 10

I am trying to round integers in python. I looked at the built-in round() function but it seems that that rounds floats.

My goal is to round integers to the closest multiple of 10. i.e.: 5-> 10, 4-> 0, 95->100, etc.

5 and higher should round up, 4 and lower should round down.

This is the code I have that does this:

def round_int(x):
    last_dig = int(str(x)[-1])
    if last_dig >= 5:
        x += 10
    return (x/10) * 10

Is this the best way to achieve what I want to achieve? Is there a built-in function that does this? Additionally, if this is the best way, is there anything wrong with the code that I missed in testing?

Answers:

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Method 1

Actually, you could still use the round function:

>>> print round(1123.456789, -1)
1120.0

This would round to the closest multiple of 10. To 100 would be -2 as the second argument and so forth.

Method 2

round() can take ints and negative numbers for places, which round to the left of the decimal. The return value is still a float, but a simple cast fixes that:

>>> int(round(5678,-1))
5680
>>> int(round(5678,-2))
5700
>>> int(round(5678,-3))
6000

Method 3

Slightly simpler:

def round_int(x):
    return 10 * ((x + 5) // 10)

Method 4

About the round(..) function returning a float

That float (double-precision in Python) is always a perfect representation of an integer, as long as it’s in the range [-2⁵³..2⁵³]. (Pedants pay attention: it’s not two’s complement in doubles, so the range is symmetric about zero.)

See the discussion here for details.

Method 5

I wanted to do the same thing, but with 5 instead of 10, and came up with a simple function. Hope it’s useful:

def roundToFive(num):
    remaining = num % 5
    if remaining in range(0, 3):
        return num - remaining
    return num + (5 - remaining)

Method 6

if you want the algebric form and still use round for it it’s hard to get simpler than:

interval = 5
n = 4
print(round(n/interval))*interval

Method 7

This function will round either be order of magnitude (right to left) or by digits the same way that format treats floating point decimal places (left to right:

def intround(n, p):
    ''' rounds an intger. if "p"<0, p is a exponent of 10; if p>0, left to right digits '''
    if p==0: return n
    if p>0:  
        ln=len(str(n))  
        p=p-ln+1 if n<0 else p-ln
    return (n + 5 * 10**(-p-1)) // 10**-p * 10**-p

>>> tgt=5555555
>>> d=2
>>> print('t{} rounded to {} places:nt{} right to left nt{} left to right'.format(
        tgt,d,intround(tgt,-d), intround(tgt,d)))

Prints

5555555 rounded to 2 places:
5555600 right to left 
5600000 left to right

You can also use Decimal class:

import decimal
import sys

def ri(i, prec=6):
    ic=long if sys.version_info.major<3 else int
    with decimal.localcontext() as lct:
        if prec>0:
            lct.prec=prec
        else:
            lct.prec=len(str(decimal.Decimal(i)))+prec  
        n=ic(decimal.Decimal(i)+decimal.Decimal('0'))
    return n

On Python 3 you can reliably use round with negative places and get a rounded integer:

def intround2(n, p):
    ''' will fail with larger floating point numbers on Py2 and require a cast to an int '''
    if p>0:
        return round(n, p-len(str(n))+1)
    else:
        return round(n, p)

On Python 2, round will fail to return a proper rounder integer on larger numbers because round always returns a float:

>>> round(2**34, -5)
17179900000.0                     # OK
>>> round(2**64, -5)
1.84467440737096e+19              # wrong

The other 2 functions work on Python 2 and 3

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0