Round integer in python without using round()

Question

I know that I can round to the closest multiple of 10 in python by using round(<int>, -1) Is there a way to do this without using this built in function?

Edit:

Thanks for the feedback! The answer using divmod() was interesting because I had never used divmod before. Just in case anyone wants to know on CodingBat the solution used modulo as was suggested in the comments. Here it is in case anyone is interested.

def round10(num):
  mod = num % 10
  num -= mod
  if mod >= 5: num += 10
  return num

It's not, it is from a practice puzzle on CodingBat.com I was just trying to do it without using the round() function. The puzzle is here codingbat.com/prob/p179960 — Manny_G
– Manny_G, Commented Dec 26, 2013 at 18:37

mgilson · Accepted Answer · 2013-12-26 18:46:57Z

2

divide by 10, int, multiply by 10.

Actually, you can do it without any builtins using the // operator:

>>> def round(x):
...     return (x//10)*10
... 
>>> round(15.)
10.0
>>> round(25.)
20.0

Of course, this always rounds down. If you want to round up for values with a remainder greater than 5, you could use divmod:

def round(x):
    n, remainder = divmod(x, 10)
    if remainder >= 5:
        n += 1
    return n * 10

edited Dec 26, 2013 at 18:46

answered Dec 26, 2013 at 18:34

mgilson

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3 Comments

DSM Over a year ago

This won't round to the "closest multiple of 10", though (at least not how I'd interpret that phrase.)

Jason Sperske Over a year ago

This is one implementation of rounding (round down) but how would you deal with the more common (in the US at least) of round half up?

mgilson Over a year ago

@JasonSperske -- Added that too.

salezica · Accepted Answer · 2013-12-26 18:43:22Z

2

This rounds to the closest, not always up or always down:

def xround(n):
    return (n + 5) // 10 * 10

answered Dec 26, 2013 at 18:43

salezica

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1 Comment

mgilson Over a year ago

This is clever. I've seen it used before -- Not sure why I didn't think of it. I would suggest integer division for python 3 compatability though (and for use with floats).

Robert Kajic · Accepted Answer · 2013-12-26 18:46:18Z

0

Round down to the closest multiple of 10:

int(n / 10) * 10

Round up to the closest multiple of 10:

int((n + 10) / 10) * 10

Round to the closest multiple of ten:

(int(n / 10) + bool(n % 10 >= 5)) * 10

edited Dec 26, 2013 at 18:46

answered Dec 26, 2013 at 18:35

Robert Kajic

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1 Comment

Robert Kajic Over a year ago

I would appreciate it if whoever downvoted would give a reason for doing so, so that I can improve my answer.

Cold Diamondz · Accepted Answer · 2015-02-21 18:17:11Z

0

I have a function that rounds n to the nearest multiple of d:

def cm(n,d,o="+"):
    p = d - (n % d)
    m = n % d
    nm = n - m
    np = n + p
    if p > m:
        return nm
    elif m > p:
        return np
    else:
        if o == "+":
            return np
        else:
            return nm

How to use: use cm(number, near multiple wanted, preferred direction in special cases)

Examples:

cm(8,10) = 10
cm(6,4,"-") = 4 #special case (n % d == d - n % d)
cm(6,4,"+") = 8 #special case
cm(6,4) = 8

answered Feb 21, 2015 at 18:17

Cold Diamondz

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