Round to 5 (or other number) in Python

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Question :

Round to 5 (or other number) in Python

Is there a built-in function that can round like the following?

10 -> 10
12 -> 10
13 -> 15
14 -> 15
16 -> 15
18 -> 20
Asked By: Pydev UA

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Answer #1:

I don’t know of a standard function in Python, but this works for me:

Python 2

def myround(x, base=5):
    return int(base * round(float(x)/base))

Python3

def myround(x, base=5):
    return base * round(x/base)

It is easy to see why the above works. You want to make sure that your number divided by 5 is an integer, correctly rounded. So, we first do exactly that (round(float(x)/5) where float is only needed in Python2), and then since we divided by 5, we multiply by 5 as well. The final conversion to int is because round() returns a floating-point value in Python 2.

I made the function more generic by giving it a base parameter, defaulting to 5.

Answered By: Alok Singhal

Answer #2:

For rounding to non-integer values, such as 0.05:

def myround(x, prec=2, base=.05):
  return round(base * round(float(x)/base),prec)

I found this useful since I could just do a search and replace in my code to change “round(” to “myround(“, without having to change the parameter values.

Answered By: CCKx

Answer #3:

It’s just a matter of scaling

>>> a=[10,11,12,13,14,15,16,17,18,19,20]
>>> for b in a:
...     int(round(b/5.0)*5.0)
... 
10
10
10
15
15
15
15
15
20
20
20
Answered By: amo-ej1

Answer #4:

Removing the ‘rest’ would work:

rounded = int(val) - int(val) % 5

If the value is aready an integer:

rounded = val - val % 5

As a function:

def roundint(value, base=5):
    return int(value) - int(value) % int(base)
Answered By: hgdeoro

Answer #5:

def round_to_next5(n):
    return n + (5 - n) % 5
Answered By: Andy Wong

Answer #6:

round(x[, n]): values are rounded to the closest multiple of 10 to the power minus n. So if n is negative…

def round5(x):
    return int(round(x*2, -1)) / 2

Since 10 = 5 * 2, you can use integer division and multiplication with 2, rather than float division and multiplication with 5.0. Not that that matters much, unless you like bit shifting

def round5(x):
    return int(round(x << 1, -1)) >> 1
Answered By: pwdyson

Answer #7:

Sorry, I wanted to comment on Alok Singhai’s answer, but it won’t let me due to a lack of reputation =/

Anyway, we can generalize one more step and go:

def myround(x, base=5):
    return base * round(float(x) / base)

This allows us to use non-integer bases, like .25 or any other fractional base.

Answered By: Aku

Answer #8:

Use:

>>> def round_to_nearest(n, m):
        r = n % m
        return n + m - r if r + r >= m else n - r

It does not use multiplication and will not convert from/to floats.

Rounding to the nearest multiple of 10:

>>> for n in range(-21, 30, 3): print('{:3d}  =>  {:3d}'.format(n, round_to_nearest(n, 10)))
-21  =>  -20
-18  =>  -20
-15  =>  -10
-12  =>  -10
 -9  =>  -10
 -6  =>  -10
 -3  =>    0
  0  =>    0
  3  =>    0
  6  =>   10
  9  =>   10
 12  =>   10
 15  =>   20
 18  =>   20
 21  =>   20
 24  =>   20
 27  =>   30

As you can see, it works for both negative and positive numbers. Ties (e.g. -15 and 15) will always be rounded upwards.

A similar example that rounds to the nearest multiple of 5, demonstrating that it also behaves as expected for a different “base”:

>>> for n in range(-21, 30, 3): print('{:3d}  =>  {:3d}'.format(n, round_to_nearest(n, 5)))
-21  =>  -20
-18  =>  -20
-15  =>  -15
-12  =>  -10
 -9  =>  -10
 -6  =>   -5
 -3  =>   -5
  0  =>    0
  3  =>    5
  6  =>    5
  9  =>   10
 12  =>   10
 15  =>   15
 18  =>   20
 21  =>   20
 24  =>   25
 27  =>   25
Answered By: wouter bolsterlee

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