Nicely representing a floating-point number in python [duplicate]

Question

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 need to be able to specify the number of significant digits.
• The number of significant digits needs to be variable, which can't be done with with the string formatting operator. [edit] I've been corrected; the string formatting operator can do this.
• I need it to be rounded the way a person would expect, not something like 1.999999999999

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?

Casia

Are there any languages that do have a built-in function for this? – Ancy

Just in case: you do know the '%.*g' % (3, 3.142596) formatting option? (Especially the asterisk) – Platinum

I did miss the asterisk formatting option, however using 'g' returns exponents and using 'f' is far from perfect: "%.*f" % (0, 34500000000000000000000) returns '34499999999999999000000' – Casia

You don't say if this is what you're doing, but if you're using floating-point math to do calculations related to money, it is a terrible, terrible idea. Either use the decimal module or store the quantity as an integer cents (or the local currency equivalent). Decimal is really your best bet. – Hymettus

Answer 1

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):
    # https://mcmap.net/q/374268/-nicely-representing-a-floating-point-number-in-python-duplicate/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)

Faddist

That's some beautiful code. It's exactly what I wanted. If you change it so that it takes a Decimal instead of float (which is trivial to do), then we can avoid floating-point errors all together. The only problem I find is with long integers. For example, try (34500000000000000000000, 17, '34500000000000000000000'). Though, again, this is probably just from floating-point errors. – Casia

Good catch. I made a slight change to the definition of d which fixes the problem with long ints, and as a side-effect, allows you to also pass numeric strings for the number arg as well. – Faddist

That fix works, but I found other, unrelated errors. Using (756867, 7, '756867.0'), I get back 75686.7. It seems the conditions for padding the result are a little off. – Casia

Answer 2

If you want floating point precision you need to use the decimal module, which is part of the Python Standard Library:

>>> import decimal
>>> d = decimal.Decimal('0.0000000000368568')
>>> print '%.15f' % d
0.000000000036857

Cheroot

The decimal module looks like it might help, but this really doesn't answer my question. – Casia

@dln385: How so? It meets all the requirements you listed. – Mickens

1. That specifies precision, not significant digits. There's a big difference. 2. It won't round past the decimal point. For example, 756867 with 3 significant digits is 757000. See the original question. 3. That method breaks down with large numbers, such as long ints. – Casia

@dln385: Did you read the docs? 1. "The decimal module incorporates a notion of significant places so that 1.30 + 1.20 is 2.50. The trailing zero is kept to indicate significance. This is the customary presentation for monetary applications. For multiplication, the “schoolbook” approach uses all the figures in the multiplicands. For instance, 1.3 * 1.2 gives 1.56 while 1.30 * 1.20 gives 1.5600." 2. it uses fixed point rather than floating so you don't run into such issues in the first place and 3. the decimal module supports arbitrary precision. – Mickens

@Billy ONeal: Yes, I did read the docs. In my last comment, I was explaining why '%.15f' % d didn't answer my question. As for the decimal module in general, I could not find a method that rounds to significant digits (including to the left of the decimal point) or displays the number without exponents. Also, it should be noted that '%.15f' % d still breaks down for large integers, even if decimals are being used. – Casia

Answer 3

Here is a snippet that formats a value according to the given error bars.

from math import floor, log10, round

def sigfig3(v, errplus, errmin):
    i = int(floor(-log10(max(errplus,errmin)) + 2))
    if i > 0:
        fmt = "%%.%df" % (i)
        return "{%s}^{%s}_{%s}" % (fmt % v,fmt % errplus, fmt % errmin)
    else:
        return "{%d}^{%d}_{%d}" % (round(v, i),round(errplus, i), numpy.round(i))

Examples:

5268685 (+1463262,-2401422) becomes 5300000 (+1500000,-2400000)
0.84312 +- 0.173124 becomes 0.84 +- 0.17

Sheldonshelduck

Answer 4

Arbitrary precision floats are needed to properly answer this question. Therefore using the decimal module is a must. There is no method to convert a decimal to a string without ever using the exponential format (part of the original question), so I wrote a function to do just that:

def removeExponent(decimal):
    digits = [str(n) for n in decimal.as_tuple().digits]
    length = len(digits)
    exponent = decimal.as_tuple().exponent
    if length <= -1 * exponent:
        zeros = -1 * exponent - length
        digits[0:0] = ["0."] + ["0"] * zeros
    elif 0 < -1 * exponent < length:
        digits.insert(exponent, ".")
    elif 0 <= exponent:
        digits.extend(["0"] * exponent)
    sign = []
    if decimal.as_tuple().sign == 1:
        sign = ["-"]
    print "".join(sign + digits)

The problem is trying to round to significant figures. Decimal's "quantize()" method won't round higher than the decimal point, and the "round()" function always returns a float. I don't know if these are bugs, but it means that the only way to round infinite precision floating point numbers is to parse it as a list or string and do the rounding manually. In other words, there is no sane answer to this question.

Casia

You don't need arbitrary precision floats. Notice that he never neds the rounded answer as a float, only as a string. BTW -1 * anything can be written as -anything. – Vermiform

I don't understand your "won't round higher than the decimal point". Try: Decimal('123.456').quantize(Decimal('1e1')), for example. – Jeri

BTW, rounding a Decimal instance returns another Decimal in Python 3.x. – Jeri

@Mark Dickinson: You're right, I didn't expect 'Decimal('123.456').quantize(Decimal('10'))' to have different behavior. Also, if Decimal rounding returns a Decimal in Python 3.x, then it seems it's time for me to upgrade. Thanks for your pointers. – Casia