I want to represent a floating-point number as a string rounded to some number of significant digits, and never using the exponential format. Essentially, I want to display any floating-point number and make sure it “looks nice”.
There are several parts to this problem:
I've figured out one way of doing this, though it looks like a work-round and it's not quite perfect. (The maximum precision is 15 significant digits.)
>>> def f(number, sigfig):
return ("%.15f" % (round(number, int(-1 * floor(log10(number)) + (sigfig - 1))))).rstrip("0").rstrip(".")
>>> print f(0.1, 1)
0.1
>>> print f(0.0000000000368568, 2)
0.000000000037
>>> print f(756867, 3)
757000
Is there a better way to do this? Why doesn't Python have a built-in function for this?
The float type in Python represents the floating point number. Float is used to represent real numbers and is written with a decimal point dividing the integer and fractional parts. For example, 97.98, 32.3+e18, -32.54e100 all are floating point numbers.
Python's built-in float type has double precision (it's a C double in CPython, a Java double in Jython). If you need more precision, get NumPy and use its numpy. float128 . 0.1 + 0.2 is not exact 0.3 in python, in every other language this is a float problem but never a double problem.
Floating-Point Number Representation. A floating-point number is typically expressed in the scientific notation, with a fraction ( F ), and an exponent ( E ) of a certain radix ( r ), in the form of F×r^E . Decimal numbers use radix of 10 ( F×10^E ); while binary numbers use radix of 2 ( F×2^E ).
Use String Formatting to Round a Number to the Given Significant Digit in Python. In Python, the %g specifier in string formats a float rounded to a specified significant figure. If the magnitude of the final number is huge, then it shows it in scientific notation. This method returns a string.
It appears there is no built-in string formatting trick which allows you to (1) print floats whose first significant digit appears after the 15th decimal place and (2) not in scientific notation. So that leaves manual string manipulation.
Below I use the decimal
module to extract the decimal digits from the float.
The float_to_decimal
function is used to convert the float to a Decimal
object. The obvious way decimal.Decimal(str(f))
is wrong because str(f)
can lose significant digits.
float_to_decimal
was lifted from the decimal module's documentation.
Once the decimal digits are obtained as a tuple of ints, the code below does the obvious thing: chop off the desired number of sigificant digits, round up if necessary, join the digits together into a string, tack on a sign, place a decimal point and zeros to the left or right as appropriate.
At the bottom you'll find a few cases I used to test the f
function.
import decimal
def float_to_decimal(f):
# http://docs.python.org/library/decimal.html#decimal-faq
"Convert a floating point number to a Decimal with no loss of information"
n, d = f.as_integer_ratio()
numerator, denominator = decimal.Decimal(n), decimal.Decimal(d)
ctx = decimal.Context(prec=60)
result = ctx.divide(numerator, denominator)
while ctx.flags[decimal.Inexact]:
ctx.flags[decimal.Inexact] = False
ctx.prec *= 2
result = ctx.divide(numerator, denominator)
return result
def f(number, sigfig):
# http://stackoverflow.com/questions/2663612/nicely-representing-a-floating-point-number-in-python/2663623#2663623
assert(sigfig>0)
try:
d=decimal.Decimal(number)
except TypeError:
d=float_to_decimal(float(number))
sign,digits,exponent=d.as_tuple()
if len(digits) < sigfig:
digits = list(digits)
digits.extend([0] * (sigfig - len(digits)))
shift=d.adjusted()
result=int(''.join(map(str,digits[:sigfig])))
# Round the result
if len(digits)>sigfig and digits[sigfig]>=5: result+=1
result=list(str(result))
# Rounding can change the length of result
# If so, adjust shift
shift+=len(result)-sigfig
# reset len of result to sigfig
result=result[:sigfig]
if shift >= sigfig-1:
# Tack more zeros on the end
result+=['0']*(shift-sigfig+1)
elif 0<=shift:
# Place the decimal point in between digits
result.insert(shift+1,'.')
else:
# Tack zeros on the front
assert(shift<0)
result=['0.']+['0']*(-shift-1)+result
if sign:
result.insert(0,'-')
return ''.join(result)
if __name__=='__main__':
tests=[
(0.1, 1, '0.1'),
(0.0000000000368568, 2,'0.000000000037'),
(0.00000000000000000000368568, 2,'0.0000000000000000000037'),
(756867, 3, '757000'),
(-756867, 3, '-757000'),
(-756867, 1, '-800000'),
(0.0999999999999,1,'0.1'),
(0.00999999999999,1,'0.01'),
(0.00999999999999,2,'0.010'),
(0.0099,2,'0.0099'),
(1.999999999999,1,'2'),
(1.999999999999,2,'2.0'),
(34500000000000000000000, 17, '34500000000000000000000'),
('34500000000000000000000', 17, '34500000000000000000000'),
(756867, 7, '756867.0'),
]
for number,sigfig,answer in tests:
try:
result=f(number,sigfig)
assert(result==answer)
print(result)
except AssertionError:
print('Error',number,sigfig,result,answer)
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With