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

Python 3

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(x/5)), and then since we divided by 5, we multiply by 5 as well.

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

Python 2

In Python 2, float(x) would be needed to ensure that / does floating-point division, and a final conversion to int is needed because round() returns a floating-point value in Python 2.

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

This should work: round(x*2)/2

Since you mention numpy

np.around(df.A.values/5, decimals=0)*5
Out[31]: array([505., 510., 610., 615.])

You can use a combination of math.ceil and simple arithmetic operations to get where you need:

import math
a = 2.3
rounded_a = math.ceil(a * 2) / 2
print(rounded_a) # 2.5

Python's builtin round function can be used to round a number to a certain number of decimal places. By multiplying the number by 2, rounding it to no decimal places and dividing it again, we can round a number to the nearest 0.5:

num = 3.46
result = round(num * 2) / 2
print(result)  # 3.5

As a function this is:

def to_nearest_half(num):
    return round(num * 2) / 2

The function can also be generalised to any decimal:

def to_nearest(num, decimal):
   return round(num / decimal) * decimal

The math.ceil (ceiling) function returns the smallest integer higher or equal to x.

For Python 3:

import math
print(math.ceil(4.2))

For Python 2:

import math
print(int(math.ceil(4.2)))