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

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)

Answer 2

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.

Answer 3

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

Answer 4

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

Answer 5

You can also use:

import math

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

In Python 3.6 it would work.

Answer 6

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)]