Python - Round to nearest 05

Question

How can I do the following rounding in python:

Round to the nearest 0.05 decimal

7,97 -> 7,95

6,72 -> 6,70

31,06 -> 31,05

36,04 -> 36,05

5,25 -> 5,25

Hope it makes sense.

Solution 1

def round_to(n, precision):
    correction = 0.5 if n >= 0 else -0.5
    return int( n/precision+correction ) * precision

def round_to_05(n):
    return round_to(n, 0.05)

Solution 2

def round05(number):
    return (round(number * 20) / 20)

Or more generically:

def round_to_value(number,roundto):
    return (round(number / roundto) * roundto)

The only problem is because you're using floats you won't get exactly the answers you want:

>>> round_to_value(36.04,0.05)
36.050000000000004

Solution 3

Using lambda function:

>>> nearest_half = lambda x: round(x * 2) / 2
>>> nearest_half(5.2)
5.0
>>> nearest_half(5.25)
5.5
>>> nearest_half(5.26)
5.5
>>> nearest_half(5.5)
5.5
>>> nearest_half(5.75)
6.0

Solution 4

Here's a one liner

def roundto(number, multiple):
   return number+multiple/2 - ((number+multiple/2) % multiple)

Solution 5

There we go.

round(VALUE*2.0, 1) / 2.0

regards

View more solutions

Comments

[martineau]

I'm surprised none of the answers that use the magic-number 20 bother to explain why it was chosen.

[Chris Morgan]

@martineau: I'll put it here for the record in case someone can't work it out. 20 == 1 / 0.05

[martineau]

decimal is the name of a module in the python standard library, so you might want to avoid using that name.

[David Webb]

round_to_05(-1) gives -0.95 which doesn't seem like the right result to me.

[David Webb]

@martineau - You're right and it's also a bad name as the second argument doesn't have to be a decimal, you can use the function to round to whole number too, e.g. round_to_value(36.04,5) gives 35.0.

[martineau]

+1 because this is currently the only answer that provides a generalized solution (which is, not surprisingly, generally better IMHO).

[fortran]

true, I was thinking in natural numbers... I'll fix it.

[John Machin]

Using "decimal" when you mean "fraction" is definitely not an aid to understanding.

[fortran]

that would round to halves, you should multiply and divide by 20

[mo-han]

correction is not needed if we use round() to replace int() like this round(n / precision) * precision. Then the wrapper func is not needed as well since this is really a simple line of code. However it does have some flaw, such as round(0.275/.05)*.05 == 0.30000000000000004. To avoid that, maybe we could use len(str(.05).split('.')[-1]) to get the number of decimal digits and use round() again to change 0.30000000000000004 to 0.30.