python 2.7: round a float up to next even number

Question

I would like to round up a float to the next even number.

Steps:

1) check if a number is odd or even

2) if odd, round up to next even number

I have step 1 ready, a function which checks if a give number is even or not:

def is_even(num):
    if int(float(num) * 10) % 2 == 0:
        return "True"
    else:
        return "False"

but I'm struggling with step 2....

Any advice?

Note: all floats will be positive.

Solution 1

There is no need for step 1. Just divide the value by 2, round up to the nearest integer, then multiply by 2 again:

import math

def round_up_to_even(f):
    return math.ceil(f / 2.) * 2

Demo:

>>> import math
>>> def round_up_to_even(f):
...     return math.ceil(f / 2.) * 2
... 
>>> round_up_to_even(1.25)
2
>>> round_up_to_even(3)
4
>>> round_up_to_even(2.25)
4

Solution 2

a = 3.5654
b = 2.568

a = int(a) if ((int(a) % 2) == 0) else int(a) + 1
b = int(b) if ((int(b) % 2) == 0) else int(b) + 1

print a
print b

value of a after execution

a = 4

value of b after execution

b = 2

Comments

[Donald Duck]

While this code may answer the question, providing additional context regarding how and/or why it solves the problem would improve the answer's long-term value.

[deweydb]

gives 2.0 and 4.0 etc. i think this changed in python 3 vs 2

[Martijn Pieters]

@deweydb yes, you are correct. Python 3 changed the returned type for math.ceil(); in Python 2 it returns a float with integral value, in Python 3 an int is returned.

[AleAve81]

I think it is quite obvious why this answer the problem and I like the simplicity of it, upvoted.

[Mark Dickinson]

What do you think round_to_ceil_even(2.3) should give? Right now, your function simply returns 2.3, which is definitely not an "even number". What about an input of 3.3? (Your function gives 4.3 as an output. I doubt that's what the OP wanted.)

[Néstor Waldyd]

Function updated.