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 would give 2.13 which is not what I need)

The function np.round accepts a decimals parameter but it appears that the functions ceil and floor don’t accept a number of decimals and always return a number with zero decimals.

Of course I can multiply the number by 10^ndecimals, then apply floor and finally divide by 10^ndecimals

new_value = np.floor(old_value * 10**ndecimals) / 10**ndecimals

But I’m wondering if there’s a built-in function that does this without having to do the operations.

Asked By: Sembei Norimaki

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Source

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

Answered By: DjaouadNM

Answer 2

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)

Answered By: theShashankB

Answer 3

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

One hint though is to reuse round instead of multiplication + 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)

UPD:
As pointed out by @aschipfl, for whole values np.round will round to the nearest even, which will lead to unexpected results, e.g. my_ceil(11) will return 12. Here is an updated solution, free of this problem:

def my_ceil(a, precision=0):
    return np.true_divide(np.ceil(a * 10**precision), 10**precision)

def my_floor(a, precision=0):
    return np.true_divide(np.floor(a * 10**precision), 10**precision)

Answered By: Marat

Answer 4

For C++ devs or a general solution that worked for me:

float a, b;

// a = 2.05
// b = 2

c = (ceil((a-b)*1000))/1000;

Answered By: The Flash

floor and ceil with number of decimals

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