Fastest Way to Round to the Nearest 5/100ths

Question

I have numbers that I want to go from:

1.215145156155 => 1.2
1.368161685161 => 1.35
1.578414616868 => 1.6

(*Note: the hundredths place should not be marked if it is zero.)

What's the fastest way to do this?

This is what I have right now, and it is not fast enough:

def rounder(v):
    v = str(round(float(v),2))
    if len(v) == 3: v = v + str(0)
    d0 = int(v[0])#ones
    d1 = int(v[2])#tenths
    d2 = int(v[3])#hundredths
    if d2 <= 4:
        return str(d0)+'.'+str(d1)
    elif 4 < d2 < 7:
        return str(d0)+'.'+str(d1)+str(5)
    elif d2 >= 7:
        if d1 != 9:
            return str(d0)+'.'+str(d1+1)
        if d1 == 9:
            return str(d0+1)+'.'+str(0)

Answer 1

Scale, round, unscale.

round(20*v)/20

I should warn you that the behaviour might surprise you:

>>> round(20*1.368161685161)/20
1.3500000000000001

The rounding is working correctly, but IEEE numbers can't represent 1.35 exactly. Python 2.7 is smarter about this and will choose the simplest representation, 1.35, when printing the number. The actual stored value is identical in 2.7 and earlier versions.

Answer 2

I would try

round(v * 2, 1) / 2

Should be pretty fast. The other suggested variant

round(v * 20) / 20

seems to be even slightly faster:

$ python -mtimeit "round(1.368161685161 * 2, 1) / 2"
1000000 loops, best of 3: 0.291 usec per loop
$ python -mtimeit "round(1.368161685161 * 20) / 20"
1000000 loops, best of 3: 0.248 usec per loop