Reputation: 22549

Truncate to three decimals in Python

How do I get 1324343032.324?

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

>>1324343032.324325235 * 1000 / 1000
1324343032.3243253
>>int(1324343032.324325235 * 1000) / 1000.0
1324343032.3239999
>>round(int(1324343032.324325235 * 1000) / 1000.0,3)
1324343032.3239999
>>str(1324343032.3239999)
'1324343032.32'

Upvotes: 99

Answers (21)

Abhranil Das

Reputation: 5916

You can use an additional float() around it if you want to preserve it as a float.

val = '%.3f'%(1324343032.324325235)

Upvotes: 95

Erwin Mayer

Reputation: 18670

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

import math
def truncate(number, digits) -> float:
    # Improve accuracy with floating point operations, to avoid truncate(16.4, 2) = 16.39 or truncate(-1.13, 2) = -1.12
    nbDecimals = len(str(number).split('.')[1]) 
    if nbDecimals <= digits:
        return number
    stepper = 10.0 ** digits
    return math.trunc(stepper * number) / stepper

Usage:

>>> truncate(1324343032.324325235, 3)
1324343032.324

Upvotes: 90

darklow

Reputation: 2329

Based on @solveMe asnwer (https://stackoverflow.com/a/39165933/641263) which I think is one of the most correct ways by utilising decimal context, I created following method which does the job exactly as needed:

import decimal

def truncate_decimal(dec: Decimal, digits: int) -> decimal.Decimal:
    round_down_ctx = decimal.getcontext()
    round_down_ctx.rounding = decimal.ROUND_DOWN
    new_dec = round_down_ctx.create_decimal(dec)
    return round(new_dec, digits)

Upvotes: 1

jpdeveloper

Reputation: 9

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

Upvotes: -1

gorandp

Reputation: 563

Function

def truncate(number: float, digits: int) -> float:
    pow10 = 10 ** digits
    return number * pow10 // 1 / pow10

Test code

f1 = 1.2666666
f2 = truncate(f1, 3)
print(f1, f2)

Output

1.2666666 1.266

Explain

It shifts f1 numbers digits times to the left, then cuts all decimals and finally shifts back the numbers digits times to the right.

Example in a sequence:

1.2666666 # number
1266.6666 # number * pow10
1266.0    # number * pow10 // 1
1.266     # number * pow10 // 1 / pow10

Upvotes: 6

revliscano

Reputation: 2272

Okay, this is just another approach to solve this working on the number as a string and performing a simple slice of it. This gives you a truncated output of the number instead of a rounded one.

num = str(1324343032.324325235)
i = num.index(".")
truncated = num[:i + 4]
    
print(truncated)

Output:

'1324343032.324'

Of course then you can parse:

float(truncated)

Upvotes: 2

Amit Baderia

Reputation: 4892

I am trying to generate a random number between 5 to 7 and want to limit it to 3 decimal places.

import random

num = float('%.3f' % random.uniform(5, 7))
print (num)

Upvotes: -1

Danylkaaa

Reputation: 161

I suggest next solution:

def my_floor(num, precision):
   return f'{num:.{precision+1}f}'[:-1]

my_floor(1.026456,2) # 1.02

Upvotes: 5

Chenna V

Reputation: 10523

Maybe python changed since this question, all of the below seem to work well

Python2.7

int(1324343032.324325235 * 1000) / 1000.0
float(int(1324343032.324325235 * 1000)) / 1000
round(int(1324343032.324325235 * 1000) / 1000.0,3)
# result for all of the above is 1324343032.324

Upvotes: 0

Vladimir Prudnikov

Reputation: 7242

I think the best and proper way is to use decimal module.

import decimal

a = 1324343032.324325235

decimal_val = decimal.Decimal(str(a)).quantize(
   decimal.Decimal('.001'), 
   rounding=decimal.ROUND_DOWN
)
float_val = float(decimal_val)

print(decimal_val)
>>>1324343032.324

print(float_val)
>>>1324343032.324

You can use different values for rounding=decimal.ROUND_DOWN, available options are ROUND_CEILING, ROUND_DOWN, ROUND_FLOOR, ROUND_HALF_DOWN, ROUND_HALF_EVEN, ROUND_HALF_UP, ROUND_UP, and ROUND_05UP. You can find explanation of each option here in docs.

Upvotes: 4

FFT

Reputation: 969

a = 1.0123456789
dec = 3 # keep this many decimals
p = 10 # raise 10 to this power
a * 10 ** p // 10 ** (p - dec) / 10 ** dec
>>> 1.012

Upvotes: 0

alsjflalasjf.dev

Reputation: 1709

'%.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.

Upvotes: 7

OSchubert

Reputation: 41

Maybe this way:

def myTrunc(theNumber, theDigits):

    myDigits = 10 ** theDigits
    return (int(theNumber * myDigits) / myDigits)

Upvotes: 4

Yiqing Lan

Reputation: 111

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'

Upvotes: 6

Squid0ne

Reputation: 11

You can also use:

import math

nValeur = format(float(input('Quelle valeur ?    ')), '.3f')

In Python 3.6 it would work.

Upvotes: 0

Eduardo Alvim

Reputation: 341

After looking for a way to solve this problem, without loading any Python 3 module or extra mathematical operations, I solved the problem using only str.format() e .float(). I think this way is faster than using other mathematical operations, like in the most commom solution. I needed a fast solution because I work with a very very large dataset and so for its working very well here.

def truncate_number(f_number, n_decimals):
      strFormNum = "{0:." + str(n_decimals+5) + "f}"
      trunc_num = float(strFormNum.format(f_number)[:-5])
      return(trunc_num)

# Testing the 'trunc_num()' function
test_num = 1150/252
[(idx, truncate_number(test_num, idx)) for idx in range(0, 20)]

It returns the following output:

[(0, 4.0),
 (1, 4.5),
 (2, 4.56),
 (3, 4.563),
 (4, 4.5634),
 (5, 4.56349),
 (6, 4.563492),
 (7, 4.563492),
 (8, 4.56349206),
 (9, 4.563492063),
 (10, 4.5634920634),
 (11, 4.56349206349),
 (12, 4.563492063492),
 (13, 4.563492063492),
 (14, 4.56349206349206),
 (15, 4.563492063492063),
 (16, 4.563492063492063),
 (17, 4.563492063492063),
 (18, 4.563492063492063),
 (19, 4.563492063492063)]

Upvotes: 1

Mak Meriales

Reputation: 1

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

1324343032.3243253

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

1324343032.324

Upvotes: -1

solveMe

Reputation: 2101

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

Upvotes: 36

AlG

Reputation: 15167

How about this:

In [1]: '%.3f' % round(1324343032.324325235 * 1000 / 1000,3)
Out[1]: '1324343032.324'

Possible duplicate of round() in Python doesn't seem to be rounding properly

[EDIT]

Given the additional comments I believe you'll want to do:

In : Decimal('%.3f' % (1324343032.324325235 * 1000 / 1000))
Out: Decimal('1324343032.324')

The floating point accuracy isn't going to be what you want:

In : 3.324
Out: 3.3239999999999998

(all examples are with Python 2.6.5)

Upvotes: 14

primelens

Reputation: 457

Use the decimal module. But if you must use floats and still somehow coerce them into a given number of decimal points converting to string an back provides a (rather clumsy, I'm afraid) method of doing it.

>>> q = 1324343032.324325235 * 1000 / 1000
>>> a = "%.3f" % q
>>> a
'1324343032.324'
>>> b = float(a)
>>> b
1324343032.324

So:

float("%3.f" % q)

Upvotes: 6

Spencer Rathbun

Reputation: 14910

Almo's link explains why this happens. To solve the problem, use the decimal library.

Upvotes: 2

Truncate to three decimals in Python

Answers (21)

Function

Test code

Output

Explain

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