round() doesn't seem to be rounding properly

swilliams_736 0
Viewed 258102

The documentation for the round() function states that you pass it a number, and the positions past the decimal to round. Thus it should do this:

n = 5.59
round(n, 1) # 5.6

But, in actuality, good old floating point weirdness creeps in and you get:

5.5999999999999996

For the purposes of UI, I need to display 5.6. I poked around the Internet and found some documentation that this is dependent on my implementation of Python. Unfortunately, this occurs on both my Windows dev machine and each Linux server I've tried. See here also.

Short of creating my own round library, is there any way around this?

floating-point rounding python
20 Answers

I can't help the way it's stored, but at least formatting works correctly: 

'%.1f' % round(n, 1) # Gives you '5.6'
Jimmy_44350

Formatting works correctly even without having to round:

"%.1f" % n
Vinko_Vrsalovic_51900

round(5.59, 1) is working fine. The problem is that 5.6 cannot be represented exactly in binary floating point.

>>> 5.6
5.5999999999999996
>>>

As Vinko says, you can use string formatting to do rounding for display.

Python has a module for decimal arithmetic if you need that.

Will_Harris_47020

You get '5.6' if you do str(round(n, 1)) instead of just round(n, 1).

dob_6160

You can switch the data type to an integer:

>>> n = 5.59
>>> int(n * 10) / 10.0
5.5
>>> int(n * 10 + 0.5)
56

And then display the number by inserting the locale's decimal separator.

However, Jimmy's answer is better.

user_3380

Floating point math is vulnerable to slight, but annoying, precision inaccuracies. If you can work with integer or fixed point, you will be guaranteed precision.

user_33470

printf the sucker.

print '%.1f' % 5.59  # returns 5.6
user_39200

I would avoid relying on round() at all in this case. Consider

print(round(61.295, 2))
print(round(1.295, 2))

will output

61.3
1.29

which is not a desired output if you need solid rounding to the nearest integer. To bypass this behavior go with math.ceil() (or math.floor() if you want to round down):

from math import ceil
decimal_count = 2
print(ceil(61.295 * 10 ** decimal_count) / 10 ** decimal_count)
print(ceil(1.295 * 10 ** decimal_count) / 10 ** decimal_count)

outputs

61.3
1.3

Hope that helps.

user_46096590

You can use the string format operator %, similar to sprintf.

mystring = "%.2f" % 5.5999
user_1570

Another potential option is:

def hard_round(number, decimal_places=0):
    """
    Function:
    - Rounds a float value to a specified number of decimal places
    - Fixes issues with floating point binary approximation rounding in python
    Requires:
    - `number`:
        - Type: int|float
        - What: The number to round
    Optional:
    - `decimal_places`:
        - Type: int 
        - What: The number of decimal places to round to
        - Default: 0
    Example:
    ```
    hard_round(5.6,1)
    ```
    """
    return int(number*(10**decimal_places)+0.5)/(10**decimal_places)
user_120141560

Here is an easy way to round a float number to any number of decimal places, and it still works in 2021!

float_number = 12.234325335563
rounded = round(float_number, 3) # 3 is the number of decimal places to be returned.You can pass any number in place of 3 depending on how many decimal places you want to return.
print(rounded)

And this will print;

12.234
user_137891350

What about:

round(n,1)+epsilon
ima_57330
Related