How to print float to n decimal places including trailing 0s?

I need to print or convert a float number to 15 decimal place string even if the result has many trailing 0s eg:

1.6 becomes 1.6000000000000000

I tried round(6.2,15) but it returns 6.2000000000000002 adding a rounding error

I also saw various people online who put the float into a string and then added trailing 0’s manually but that seems bad…

What is the best way to do this?

Answers:

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Method 1

For Python versions in 2.6+ and 3.x

You can use the str.format method. Examples:

>>> print('{0:.16f}'.format(1.6))
1.6000000000000001

>>> print('{0:.15f}'.format(1.6))
1.600000000000000

Note the 1 at the end of the first example is rounding error; it happens because exact representation of the decimal number 1.6 requires an infinite number binary digits. Since floating-point numbers have a finite number of bits, the number is rounded to a nearby, but not equal, value.

For Python versions prior to 2.6 (at least back to 2.0)

You can use the “modulo-formatting” syntax (this works for Python 2.6 and 2.7 too):

>>> print '%.16f' % 1.6
1.6000000000000001

>>> print '%.15f' % 1.6
1.600000000000000

Method 2

The cleanest way in modern Python >=3.6, is to use an f-string with string formatting:

>>> var = 1.6
>>> f"{var:.15f}"
'1.600000000000000'

Method 3

Floating point numbers lack precision to accurately represent “1.6” out to that many decimal places. The rounding errors are real. Your number is not actually 1.6.

Check out: http://docs.python.org/library/decimal.html

Method 4

We can use format() to print digits after the decimal places.
Taken from http://docs.python.org/tutorial/floatingpoint.html

>>> format(math.pi, '.12g')  # give 12 significant digits
'3.14159265359'

>>> format(math.pi, '.2f')   # give 2 digits after the point
'3.14'

Method 5

I guess this is essentially putting it in a string, but this avoids the rounding error:

import decimal

def display(x):
    digits = 15
    temp = str(decimal.Decimal(str(x) + '0' * digits))
    return temp[:temp.find('.') + digits + 1]

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0