很好地代表python中的浮点数

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?

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):
    # 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)

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

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

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.