Round up to Second Decimal Place in Python

How can I round up a number to the second decimal place in python? For example:

0.022499999999999999

Should round up to 0.03

0.1111111111111000

Should round up to 0.12

If there is any value in the third decimal place, I want it to always round up leaving me 2 values behind the decimal point.

Python includes the round() function which lets you specify the number of digits you want. From the documentation:

round(x[, n])

Return the floating point value x rounded to n digits after the decimal point. If n is omitted, it defaults to zero. The result is a floating point number. Values are rounded to the closest multiple of 10 to the power minus n; if two multiples are equally close, rounding is done away from 0 (so. for example, round(0.5) is 1.0 and round(-0.5) is -1.0).

So you would want to use round(x, 2) to do normal rounding. To ensure that the number is always rounded up you would need to use the ceil(x) function. Similarly, to round down use floor(x).

from math import ceil

num = 0.1111111111000
num = ceil(num * 100) / 100.0

See:
math.ceil documentation
round documentation - You'll probably want to check this out anyway for future reference

x = math.ceil(x * 100.0) / 100.0

Updated answer:
The problem with my original answer, as pointed out in the comments by @jpm, is the behavior at the boundaries. Python 3 makes this even more difficult since it uses "bankers" rounding instead of "old school" rounding. However, in looking into this issue I discovered an even better solution using the decimal library.

import decimal

def round_up(x, place=0):
    context = decimal.getcontext()
    # get the original setting so we can put it back when we're done
    original_rounding = context.rounding
    # change context to act like ceil()
    context.rounding = decimal.ROUND_CEILING

    rounded = round(decimal.Decimal(str(x)), place)
    context.rounding = original_rounding
    return float(rounded)

Or if you really just want a one-liner:

import decimal
decimal.getcontext().rounding = decimal.ROUND_CEILING

# here's the one-liner
float(round(decimal.Decimal(str(0.1111)), ndigits=2))
>> 0.12

# Note: this only affects the rounding of `Decimal`
round(0.1111, ndigits=2)
>> 0.11

Here are some examples:

round_up(0.022499999999999999, 2)
>> 0.03
round_up(0.1111111111111000, 2)
>> 0.12
round_up(0.1111111111111000, 3)
>> 0.112

round_up(3.4)
>> 4.0

# @jpm - boundaries do what we want
round_up(0.1, 2)
>> 0.1
round_up(1.1, 2)
>> 1.1

# Note: this still rounds toward `inf`, not "away from zero"
round_up(2.049, 2)
>> 2.05
round_up(-2.0449, 2)
>> -2.04

We can use it to round to the left of the decimal as well:

round_up(11, -1)
>> 20

We don't multiply by 10, thereby avoiding the overflow mentioned in this answer.

round_up(1.01e308, -307)
>> 1.1e+308

Original Answer (Not recommended):
This depends on the behavior you want when considering positive and negative numbers, but if you want something that always rounds to a larger value (e.g. 2.0449 -> 2.05, -2.0449 -> -2.04) then you can do:

round(x + 0.005, 2)

or a little fancier:

def round_up(x, place):
    return round(x + 5 * 10**(-1 * (place + 1)), place)

This also seems to work as follows:

round(144, -1)
# 140
round_up(144, -1)
# 150
round_up(1e308, -307)
# 1.1e308

Extrapolating from Edwin's answer:

from math import ceil, floor
def float_round(num, places = 0, direction = floor):
    return direction(num * (10**places)) / float(10**places)

To use:

>>> float_round(0.21111, 3, ceil)  #round up
>>> 0.212
>>> float_round(0.21111, 3)        #round down
>>> 0.211
>>> float_round(0.21111, 3, round) #round naturally
>>> 0.211