Comment 17 for bug 882036

Well, Python's round() seems to be an asymmetrical half up algorithm, as opposed to the symmetrical half up rounding in common math. Using epsilon will not correct that. I grand you the sign issue, but not the precision issue you mentioned.

Corrected algorithm:

def roundf(f, prec=0):
    return round(f + cmp(f, 0.0) * (.1 / pow(10, prec + 2), prec)

>>> roundf(-2.675000000000005, 14)
-2.67500000000001
>>> roundf(2.675000000000005, 14)
2.67500000000001
>>> roundf(2.675000000000005, 15)
2.675000000000005
>>> roundf((-2.675000000000005, 15)
-2.675000000000005
>>> roundf(0.575,2)
0.58
>>> roundf(2.675,2)
2.68
>>> roundf(-2.678,2)
-2.68

Cheers