Round up to Second Decimal Place in Python

Question

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

0.022499999999999999

Should round up to 0.03<

Answer 1

Note that the ceil(num * 100) / 100 trick will crash on some degenerate inputs, like 1e308. This may not come up often but I can tell you it just cost me a couple of days. To avoid this, "it would be nice if" ceil() and floor() took a decimal places argument, like round() does... Meanwhile, anyone know a clean alternative that won't crash on inputs like this? I had some hopes for the decimal package but it seems to die too:

>>> from math import ceil
>>> from decimal import Decimal, ROUND_DOWN, ROUND_UP
>>> num = 0.1111111111000
>>> ceil(num * 100) / 100
0.12
>>> float(Decimal(num).quantize(Decimal('.01'), rounding=ROUND_UP))
0.12
>>> num = 1e308
>>> ceil(num * 100) / 100
Traceback (most recent call last):
  File "<string>", line 301, in runcode
  File "<interactive input>", line 1, in <module>
OverflowError: cannot convert float infinity to integer
>>> float(Decimal(num).quantize(Decimal('.01'), rounding=ROUND_UP))
Traceback (most recent call last):
  File "<string>", line 301, in runcode
  File "<interactive input>", line 1, in <module>
decimal.InvalidOperation: [<class 'decimal.InvalidOperation'>]

Of course one might say that crashing is the only sane behavior on such inputs, but I would argue that it's not the rounding but the multiplication that's causing the problem (that's why, eg, 1e306 doesn't crash), and a cleaner implementation of the round-up-nth-place fn would avoid the multiplication hack.

Answer 2

The round funtion stated does not works for definate integers like :

a=8
round(a,3)
8.0
a=8.00
round(a,3)
8.0
a=8.000000000000000000000000
round(a,3)
8.0

but , works for :

r=400/3.0
r
133.33333333333334
round(r,3)
133.333

Morever the decimals like 2.675 are rounded as 2.67 not 2.68.
Better use the other method provided above.

Answer 3

Here's a simple way to do it that I don't see in the other answers.

To round up to the second decimal place:

>>> n = 0.022499999999999999
>>> 
>>> -(-n//.01) * .01
0.03
>>> 

Other value:

>>> n = 0.1111111111111000
>>> 
>>> -(-n//.01) * .01
0.12
>>> 

With floats there's the occasional value with some minute imprecision, which can be corrected for if you're displaying the values for instance:

>>> n = 10.1111111111111000
>>> 
>>> -(-n//0.01) * 0.01
10.120000000000001
>>> 
>>> f"{-(-n//0.01) * 0.01:.2f}"
'10.12'
>>> 

A simple roundup function with a parameter to specify precision:

>>> roundup = lambda n, p: -(-n//10**-p) * 10**-p
>>> 
>>> # Or if you want to ensure truncation using the f-string method:
>>> roundup = lambda n, p: float(f"{-(-n//10**-p) * 10**-p:.{p}f}")
>>> 
>>> roundup(0.111111111, 2)
0.12
>>> roundup(0.111111111, 3)
0.112

Answer 4

The python round function could be rounding the way not you expected.

You can be more specific about the rounding method by using Decimal.quantize

eg.

from decimal import Decimal, ROUND_HALF_UP
res = Decimal('0.25').quantize(Decimal('0.0'), rounding=ROUND_HALF_UP)
print(res) 
# prints 0.3

More reference:

https://gist.github.com/jackiekazil/6201722

Answer 5

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