Truncate to three decimals in Python

Question

How do I get 1324343032.324?

As you can see below, the following do not work:

>>1324343032.324325235 * 1000 / 1000
1324343032.3243253
>>i         


        

Answer 1

>>> float(1324343032.324325235) * float(1000) / float(1000)

1324343032.3243253

>>> round(float(1324343032.324325235) * float(1000) / float(1000), 3)

1324343032.324

Answer 2

'%.3f'%(1324343032.324325235)

It's OK just in this particular case.

Simply change the number a little bit:

1324343032.324725235

And then:

'%.3f'%(1324343032.324725235)

gives you 1324343032.325

Try this instead:

def trun_n_d(n,d):
    s=repr(n).split('.')
    if (len(s)==1):
        return int(s[0])
    return float(s[0]+'.'+s[1][:d])

Another option for trun_n_d:

def trun_n_d(n,d):
    dp = repr(n).find('.') #dot position
    if dp == -1:  
        return int(n) 
    return float(repr(n)[:dp+d+1])

Yet another option ( a oneliner one) for trun_n_d [this, assumes 'n' is a str and 'd' is an int]:

def trun_n_d(n,d):
    return (  n if not n.find('.')+1 else n[:n.find('.')+d+1]  )

trun_n_d gives you the desired output in both, Python 2.7 and Python 3.6

trun_n_d(1324343032.324325235,3) returns 1324343032.324

Likewise, trun_n_d(1324343032.324725235,3) returns 1324343032.324

---

Note 1 In Python 3.6 (and, probably, in Python 3.x) something like this, works just fine:

def trun_n_d(n,d):
    return int(n*10**d)/10**d

But, this way, the rounding ghost is always lurking around.

Note 2 In situations like this, due to python's number internals, like rounding and lack of precision, working with n as a str is way much better than using its int counterpart; you can always cast your number to a float at the end.

Answer 3

You can use the following function to truncate a number to a set number of decimals:

import math
def truncate(number, digits) -> float:
    stepper = 10.0 ** digits
    return math.trunc(stepper * number) / stepper

Usage:

>>> truncate(1324343032.324325235, 3)
1324343032.324

Answer 4

I've found another solution (it must be more efficient than "string witchcraft" workarounds):

>>> import decimal
# By default rounding setting in python is decimal.ROUND_HALF_EVEN
>>> decimal.getcontext().rounding = decimal.ROUND_DOWN
>>> c = decimal.Decimal(34.1499123)
# By default it should return 34.15 due to '99' after '34.14'
>>> round(c,2)
Decimal('34.14')
>>> float(round(c,2))
34.14
>>> print(round(c,2))
34.14

About decimals module

About rounding settings

Answer 5

I believe using the format function is a bad idea. Please see the below. It rounds the value. I use Python 3.6.

>>> '%.3f'%(1.9999999)
'2.000'

Use a regular expression instead:

>>> re.match(r'\d+.\d{3}', str(1.999999)).group(0)
'1.999'

Answer 6

I develop a good solution, I know there is much If statements, but It works! (Its only for <1 numbers)

def truncate(number, digits) -> float:
    startCounting = False
    if number < 1:
      number_str = str('{:.20f}'.format(number))
      resp = ''
      count_digits = 0
      for i in range(0, len(number_str)):
        if number_str[i] != '0' and number_str[i] != '.' and number_str[i] != ',':
          startCounting = True
        if startCounting:
          count_digits = count_digits + 1
        resp = resp + number_str[i]
        if count_digits == digits:
            break
      return resp
    else:
      return number