Converting a Python Float to a String without losing precision Converting a Python Float to a String without losing precision python python

Converting a Python Float to a String without losing precision


I'm the author of xlrd. There is so much confusion in other answers and comments to rebut in comments so I'm doing it in an answer.

@katriealex: """precision being lost in the guts of xlrd""" --- entirely unfounded and untrue. xlrd reproduces exactly the 64-bit float that's stored in the XLS file.

@katriealex: """It may be possible to modify your local xlrd installation to change the float cast""" --- I don't know why you would want to do this; you don't lose any precision by floating a 16-bit integer!!! In any case that code is used only when reading Excel 2.X files (which had an INTEGER-type cell record). The OP gives no indication that he is reading such ancient files.

@jloubert: You must be mistaken. "%.40r" % a_float is just a baroque way of getting the same answer as repr(a_float).

@EVERYBODY: You don't need to convert a float to a decimal to preserve the precision. The whole point of the repr() function is that the following is guaranteed:

float(repr(a_float)) == a_float

Python 2.X (X <= 6) repr gives a constant 17 decimal digits of precision, as that is guaranteed to reproduce the original value. Later Pythons (2.7, 3.1) give the minimal number of decimal digits that will reproduce the original value.

Python 2.6.4 (r264:75708, Oct 26 2009, 08:23:19) [MSC v.1500 32 bit (Intel)] on win32>>> f = 0.38288746115497402>>> repr(f)'0.38288746115497402'>>> float(repr(f)) == fTruePython 2.7 (r27:82525, Jul  4 2010, 09:01:59) [MSC v.1500 32 bit (Intel)] on win32>>> f = 0.38288746115497402>>> repr(f)'0.382887461154974'>>> float(repr(f)) == fTrue

So the bottom line is that if you want a string that preserves all the precision of a float object, use preserved = repr(the_float_object) ... recover the value later by float(preserved). It's that simple. No need for the decimal module.


You can use repr() to convert to a string without losing precision, then convert to a Decimal:

>>> from decimal import Decimal>>> f = 0.38288746115497402>>> d = Decimal(repr(f))>>> print d0.38288746115497402


EDIT: I am wrong. I shall leave this answer here so the rest of the thread makes sense, but it's not true. Please see John Machin's answer above. Thanks guys =).

If the above answers work that's great -- it will save you a lot of nasty hacking. However, at least on my system, they won't. You can check this with e.g.

import sysprint( "%.30f" % sys.float_info.epsilon )

That number is the smallest float that your system can distinguish from zero. Anything smaller than that may be randomly added or subtracted from any float when you perform an operation. This means that, at least on my Python setup, the precision is lost inside the guts of xlrd, and there seems to be nothing you can do without modifying it. Which is odd; I'd have expected this case to have occurred before, but apparently not!

It may be possible to modify your local xlrd installation to change the float cast. Open up site-packages\xlrd\sheet.py and go down to line 1099:

...elif rc == XL_INTEGER:                    rowx, colx, cell_attr, d = local_unpack('<HH3sH', data)                    self_put_number_cell(rowx, colx, float(d), self.fixed_BIFF2_xfindex(cell_attr, rowx, colx))...

Notice the float cast -- you could try changing that to a decimal.Decimal and see what happens.