floor and ceil with number of decimals

Question

I need to floor a float number with an specific number of decimals.

So:

2.1235 with 2 decimals --> 2.12
2.1276 with 2 decimals --> 2.12  (round         


        

Answer 1

If regular expressions are an option, you can give this a go:

import re

def truncate(n, d):
  return float(re.search('\d+\.\d{}'.format(d), str(float(n)))[0])

print(truncate(2.1235, 2))
print(truncate(2.1276, 2))

Output:

2.12
2.12

Another solution using str.split:

def truncate(n, d):
  s = str(float(n)).split('.')
  return float('{}.{}'.format(s[0], s[1][:d]))

print(truncate(2.1235, 2))
print(truncate(2.1276, 2))

Output:

2.12
2.12

Answer 2

Neither Python built-in nor numpy's version of ceil/floor support precision.

One hint though is to reuse round instead of multyplication + division (should be much faster):

def my_ceil(a, precision=0):
    return np.round(a + 0.5 * 10**(-precision), precision)

def my_floor(a, precision=0):
    return np.round(a - 0.5 * 10**(-precision), precision)

Answer 3

This seems to work (needs no import and works using the // operator which should be faster than numpy, as it simply returns the floor of the division):

a = 2.338888
n_decimals = 2
a = ((a*10**n_decimals)//1)/(10**n_decimals)