How to print float to n decimal places including trailing 0s?
Question:
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:
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
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
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]
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'
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'
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?
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
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
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]
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'
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'