Convert float to string in positional format (without scientific notation and false precision)

Question

I want to print some floating point numbers so that they're always written in decimal form (e.g. 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' 

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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.

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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'.

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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 floats is the same as with repr; and with f flag there are 6 fractional digits by default:

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

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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 floats 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.