Re: Getting fractional part from a float without using stringoperations

[srinivasan srinivas]

Yes it works for most of the cases.  But it doesn't for the following case:

>>str(abs(int(1 234567.89)-1234567.89))

'0.889999999898 '

Thanks,
Srini


----- Original Message ----
From: Tino Wildenhain <tino@wildenhai n.de>
To: srinivasan srinivas <sri_annauni@ya hoo.co.in>
Cc: Jeremiah Dodds <jeremiah.dodds @gmail.com>; python-list@python.org
Sent: Wednesday, 19 November, 2008 7:33:46 PM
Subject: Re: Getting fractionalpart from a float without using string operations

srinivasan srinivaswrote:

Yes. But it didn't give only the expected decimals.
For ex:
>  >>a = 1.23
>  >>abs(int(a) -a)
0.2299999999999 9998
>  I would like to get the result '0.23' only.

well, thats what get stored internally - there
is no way around it if you are using floating
point numbers:

>>0.23

0.2300000000000 0001

but str() handles the rounding correctly:

>>print 0.23

0.23

>>print abs(int(a) -a)

0.23

See also http://en.wikipedia.org/wiki/Floating_point
for the problems with FP figures.

Regards
Tino



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[James Harris]

Re: Getting fractional part from a float without using stringoperation s

On 20 Nov, 06:01, srinivasan srinivas <sri_anna...@ya hoo.co.inwrote:

Yes it works for most of the cases. But it doesn't for the following case:
>

>str(abs(int(12 34567.89)-1234567.89))

>
'0.889999999898 '

Well, that is 0.89 or about as near to it as the calculation can
represent. Like other numbers 0.89 evidently cannot be exactly
represented as a float in binary. The nearest number above it is

>>0.89

0.8900000000000 0001

>>>

This is not a Python issue but a result of storing numbers in floating
binary form.

BTW, please post comments below existing ones rather than above them.
It is more familiar on Usenet and, as a consequence, easier to read.

James