Python Output to File Higher Floating Point Precision

Question

The following:

>>>import struct
>>>struct.unpack('!d', ('40d11998e5035288').decode('hex'))[0]
>>>
>>>17510.3889778429

I would like to be able to print the value 17510.3889778429 to a .csv output file. For some reason, when I write this to a file, it rounds the decimal and only writes:

17510.3889778

How would I print this with higher precision?

Thanks, Ned

What version of Python are you using? This is a known issue that's already fixed in the most recent versions of Python (e.g., 2.7.3); see bugs.python.org/issue13573 — Mark Dickinson
– Mark Dickinson, Commented Jun 20, 2012 at 20:47
Incidentally, looks like stackoverflow.com/questions/8455253 is a possible duplicate. — Mark Dickinson
– Mark Dickinson, Commented Jun 20, 2012 at 20:48

Russell Borogove · Accepted Answer · 2012-06-20 18:38:19Z

3

Assuming you're using Python's csv module, the documentation says that non-string data is formatted using str(), and that seems to be where the truncation is happening. So one thing you could do is define your own class that stringizes the way you want:

class myFloat( float ):
    def __str__(self):
        return "%.12f"%self

 x = myFloat( 17510.3889778429 )
 print "%s"%x

Yields:

17510.388977842900

answered Jun 20, 2012 at 18:38

Russell Borogove

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Mark Ransom · Accepted Answer · 2012-06-20 18:36:02Z

1

You can convert the number to a string using any appropriate method and write the string to the CSV.

>>> x = 17510.3889778429
>>> s = str(x)
>>> s
'17510.3889778'
>>> s = '%17.11f' % x
>>> s
'17510.38897784290'

answered Jun 20, 2012 at 18:36

Mark Ransom

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