将浮点数转换为位置格式的字符串（没有科学记数法和错误精度）

Question

I want to print some floating point numbers so that they're always written in decimal form (eg 12345000000000000000000.0 or 0.000000000000012345 , not in scientific notation , yet I'd want to the result to have the up to ~15.7 significant figures of a IEEE 754 double, and no more.

What I want is ideally so that the result is the shortest string in positional decimal format that still results in the same value when converted to a float .

It is well-known that the repr of a float is written in scientific notation if the exponent is greater than 15, or less than -4:

>>> n = 0.000000054321654321
>>> n
5.4321654321e-08  # scientific notation

If str is used, the resulting string again is in scientific notation:

>>> str(n)
'5.4321654321e-08'

---

It has been suggested that I can use format with f flag and sufficient precision to get rid of the scientific notation:

>>> format(0.00000005, '.20f')
'0.00000005000000000000'

It works for that number, though it has some extra trailing zeroes. But then the same format fails for .1 , which gives decimal digits beyond the actual machine precision of float:

>>> format(0.1, '.20f')
'0.10000000000000000555'

And if my number is 4.5678e-20 , using .20f would still lose relative precision:

>>> format(4.5678e-20, '.20f')
'0.00000000000000000005'

Thus these approaches do not match my requirements .

---

This leads to the question: what is the easiest and also well-performing way to print arbitrary floating point number in decimal format, having the same digits as in repr(n) (or str(n) on Python 3) , but always using the decimal format, not the scientific notation.

That is, a function or operation that for example converts the float value 0.00000005 to string '0.00000005' ; 0.1 to '0.1' ; 420000000000000000.0 to '420000000000000000.0' or 420000000000000000 and formats the float value -4.5678e-5 as '-0.000045678' .

---

After the bounty period: It seems that there are at least 2 viable approaches, as Karin demonstrated that using string manipulation one can achieve significant speed boost compared to my initial algorithm on Python 2.

Thus,

• If performance is important and Python 2 compatibility is required; or if the decimal module cannot be used for some reason, then Karin's approach using string manipulation is the way to do it.
• On Python 3, my somewhat shorter code will also be faster .

Since I am primarily developing on Python 3, I will accept my own answer, and shall award Karin the bounty.

Answer 1

Unfortunately it seems that not even the new-style formatting with float.__format__ supports this. The default formatting of float s is the same as with repr ; and with f flag there are 6 fractional digits by default:

>>> format(0.0000000005, 'f')
'0.000000'

---

However there is a hack to get the desired result - not the fastest one, but relatively simple:

• first the float is converted to a string using str() or repr()
• then a new Decimal instance is created from that string.
• Decimal.__format__ supports f flag which gives the desired result, and, unlike float s it prints the actual precision instead of default precision.

Thus we can make a simple utility function float_to_str :

import decimal

# create a new context for this task
ctx = decimal.Context()

# 20 digits should be enough for everyone :D
ctx.prec = 20

def float_to_str(f):
    """
    Convert the given float to a string,
    without resorting to scientific notation
    """
    d1 = ctx.create_decimal(repr(f))
    return format(d1, 'f')

Care must be taken to not use the global decimal context, so a new context is constructed for this function. This is the fastest way; another way would be to use decimal.local_context but it would be slower, creating a new thread-local context and a context manager for each conversion.

This function now returns the string with all possible digits from mantissa, rounded to the shortest equivalent representation :

>>> float_to_str(0.1)
'0.1'
>>> float_to_str(0.00000005)
'0.00000005'
>>> float_to_str(420000000000000000.0)
'420000000000000000'
>>> float_to_str(0.000000000123123123123123123123)
'0.00000000012312312312312313'

The last result is rounded at the last digit

As @Karin noted, float_to_str(420000000000000000.0) does not strictly match the format expected; it returns 420000000000000000 without trailing .0 .

Answer 2

If you are satisfied with the precision in scientific notation, then could we just take a simple string manipulation approach? Maybe it's not terribly clever, but it seems to work (passes all of the use cases you've presented), and I think it's fairly understandable:

def float_to_str(f):
    float_string = repr(f)
    if 'e' in float_string:  # detect scientific notation
        digits, exp = float_string.split('e')
        digits = digits.replace('.', '').replace('-', '')
        exp = int(exp)
        zero_padding = '0' * (abs(int(exp)) - 1)  # minus 1 for decimal point in the sci notation
        sign = '-' if f < 0 else ''
        if exp > 0:
            float_string = '{}{}{}.0'.format(sign, digits, zero_padding)
        else:
            float_string = '{}0.{}{}'.format(sign, zero_padding, digits)
    return float_string

n = 0.000000054321654321
assert(float_to_str(n) == '0.000000054321654321')

n = 0.00000005
assert(float_to_str(n) == '0.00000005')

n = 420000000000000000.0
assert(float_to_str(n) == '420000000000000000.0')

n = 4.5678e-5
assert(float_to_str(n) == '0.000045678')

n = 1.1
assert(float_to_str(n) == '1.1')

n = -4.5678e-5
assert(float_to_str(n) == '-0.000045678')

Performance :

I was worried this approach may be too slow, so I ran timeit and compared with the OP's solution of decimal contexts. It appears the string manipulation is actually quite a bit faster. Edit : It appears to only be much faster in Python 2. In Python 3, the results were similar, but with the decimal approach slightly faster.

Result :

• Python 2: using ctx.create_decimal() : 2.43655490875

• Python 2: using string manipulation: 0.305557966232

• Python 3: using ctx.create_decimal() : 0.19519368198234588

• Python 3: using string manipulation: 0.2661344590014778

Here is the timing code:

from timeit import timeit

CODE_TO_TIME = '''
float_to_str(0.000000054321654321)
float_to_str(0.00000005)
float_to_str(420000000000000000.0)
float_to_str(4.5678e-5)
float_to_str(1.1)
float_to_str(-0.000045678)
'''
SETUP_1 = '''
import decimal

# create a new context for this task
ctx = decimal.Context()

# 20 digits should be enough for everyone :D
ctx.prec = 20

def float_to_str(f):
    """
    Convert the given float to a string,
    without resorting to scientific notation
    """
    d1 = ctx.create_decimal(repr(f))
    return format(d1, 'f')
'''
SETUP_2 = '''
def float_to_str(f):
    float_string = repr(f)
    if 'e' in float_string:  # detect scientific notation
        digits, exp = float_string.split('e')
        digits = digits.replace('.', '').replace('-', '')
        exp = int(exp)
        zero_padding = '0' * (abs(int(exp)) - 1)  # minus 1 for decimal point in the sci notation
        sign = '-' if f < 0 else ''
        if exp > 0:
            float_string = '{}{}{}.0'.format(sign, digits, zero_padding)
        else:
            float_string = '{}0.{}{}'.format(sign, zero_padding, digits)
    return float_string
'''

print(timeit(CODE_TO_TIME, setup=SETUP_1, number=10000))
print(timeit(CODE_TO_TIME, setup=SETUP_2, number=10000))

Answer 3

As of NumPy 1.14.0, you can just use numpy.format_float_positional . For example, running against the inputs from your question:

>>> numpy.format_float_positional(0.000000054321654321)
'0.000000054321654321'
>>> numpy.format_float_positional(0.00000005)
'0.00000005'
>>> numpy.format_float_positional(0.1)
'0.1'
>>> numpy.format_float_positional(4.5678e-20)
'0.000000000000000000045678'

numpy.format_float_positional uses the Dragon4 algorithm to produce the shortest decimal representation in positional format that round-trips back to the original float input. There's also numpy.format_float_scientific for scientific notation, and both functions offer optional arguments to customize things like rounding and trimming of zeros.

Answer 4

If you are ready to lose your precision arbitrary by calling str() on the float number, then it's the way to go:

import decimal

def float_to_string(number, precision=20):
    return '{0:.{prec}f}'.format(
        decimal.Context(prec=100).create_decimal(str(number)),
        prec=precision,
    ).rstrip('0').rstrip('.') or '0'

It doesn't include global variables and allows you to choose the precision yourself. Decimal precision 100 is chosen as an upper bound for str(float) length. The actual supremum is much lower. The or '0' part is for the situation with small numbers and zero precision.

Note that it still has its consequences:

>> float_to_string(0.10101010101010101010101010101)
'0.10101010101'

Otherwise, if the precision is important, format is just fine:

import decimal

def float_to_string(number, precision=20):
    return '{0:.{prec}f}'.format(
        number, prec=precision,
    ).rstrip('0').rstrip('.') or '0'

It doesn't miss the precision being lost while calling str(f) . The or

>> float_to_string(0.1, precision=10)
'0.1'
>> float_to_string(0.1)
'0.10000000000000000555'
>>float_to_string(0.1, precision=40)
'0.1000000000000000055511151231257827021182'

>>float_to_string(4.5678e-5)
'0.000045678'

>>float_to_string(4.5678e-5, precision=1)
'0'

Anyway, maximum decimal places are limited, since the float type itself has its limits and cannot express really long floats:

>> float_to_string(0.1, precision=10000)
'0.1000000000000000055511151231257827021181583404541015625'

Also, whole numbers are being formatted as-is.

>> float_to_string(100)
'100'

Answer 5

I think rstrip can get the job done.

a=5.4321654321e-08
'{0:.40f}'.format(a).rstrip("0") # float number and delete the zeros on the right
# '0.0000000543216543210000004442039220863003' # there's roundoff error though

Let me know if that works for you.

Answer 6

Interesting question, to add a little bit more of content to the question, here's a litte test comparing @Antti Haapala and @Harold solutions outputs:

import decimal
import math

ctx = decimal.Context()


def f1(number, prec=20):
    ctx.prec = prec
    return format(ctx.create_decimal(str(number)), 'f')


def f2(number, prec=20):
    return '{0:.{prec}f}'.format(
        number, prec=prec,
    ).rstrip('0').rstrip('.')

k = 2*8

for i in range(-2**8,2**8):
    if i<0:
        value = -k*math.sqrt(math.sqrt(-i))
    else:
        value = k*math.sqrt(math.sqrt(i))

    value_s = '{0:.{prec}E}'.format(value, prec=10)

    n = 10

    print ' | '.join([str(value), value_s])
    for f in [f1, f2]:
        test = [f(value, prec=p) for p in range(n)]
        print '\t{0}'.format(test)

Neither of them gives "consistent" results for all cases.

• With Anti's you'll see strings like '-000' or '000'
• With Harolds's you'll see strings like ''

I'd prefer consistency even if I'm sacrificing a little bit of speed. Depends which tradeoffs you want to assume for your use-case.

Answer 7

using format(float, ' .f '):

old = 0.00000000000000000000123
if str(old).__contains__('e-'):
    float_length = str(old)[-2:]
    new=format(old,'.'+str(float_length)+'f')
    print(old)
    print(new)