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

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