Remove integer from float number

[Derek Basch]

How can I return:

".666"

from float:

"0.666"

This is what I have so far:

"%.6f" % x

Thanks Everyone,
Derek Basch

[Larry Bates]

Derek Basch wrote:

How can I return:

".666"

from float:

"0.666"

This is what I have so far:

"%.6f" % x

Thanks Everyone,
Derek Basch

This works but I'm not entirely sure I know what you are
trying to accomplish.

("%.3f" % x)[1:]
-Larry Bates

[Derek Basch]

Ahh yes you have to put parenthases around the string formatting to
remove the integer using indexes. Thanks, that's just what I needed!

Derek Basch

[Paul Rubin]

"Derek Basch" <db****@yahoo.com> writes:

Ahh yes you have to put parenthases around the string formatting to
remove the integer using indexes. Thanks, that's just what I needed!

I think it's better to remove leading zeros explicitly:

('%.3x' % x).lstrip('0')

[John Machin]

On 24/03/2006 6:44 AM, Larry Bates wrote:

Derek Basch wrote:

How can I return:

".666"

from float:

"0.666"

This is what I have so far:

>"%.6f" % x

Thanks Everyone,
Derek Basch

This works but I'm not entirely sure I know what you are
trying to accomplish.

("%.3f" % x)[1:]

x = 12345.666; ("%.3f" % x)[1:] '2345.666'

I'm sure of neither what the OP is trying to accomplish nor what Larry's
definition of "works" is :-)

Perhaps the condition abs(x) < 1.0 is implied ...

[Larry Bates]

John Machin wrote:

On 24/03/2006 6:44 AM, Larry Bates wrote:

Derek Basch wrote:

How can I return:

".666"

from float:

"0.666"

This is what I have so far:
>> "%.6f" % x

Thanks Everyone,
Derek Basch

This works but I'm not entirely sure I know what you are
trying to accomplish.

("%.3f" % x)[1:]

x = 12345.666; ("%.3f" % x)[1:] '2345.666'

I'm sure of neither what the OP is trying to accomplish nor what Larry's
definition of "works" is :-)

Perhaps the condition abs(x) < 1.0 is implied ...

For the example given, my code works. With so little information
the only thing I could do is answer the specific question and
caveat it that I don't know "exactly" what OP is trying to accomplish.
By the OPs response to my post, it was what he was looking for.
But I agree it is very much an edge-case question.

-Larry Bates

[Arne Ludwig]

With that terse description and the subject line I would interpret the
OP like so:

print re.sub(".*\.",".","0.666") ..666 print re.sub(".*\.",".","123.666")

..666

[Grant Edwards]

On 2006-03-23, Arne Ludwig <ar**@citde.net> wrote:

With that terse description and the subject line I would interpret the
OP like so:

print re.sub(".*\.",".","0.666") .666 print re.sub(".*\.",".","123.666") .666

Or if you're allergic to regular expressions:

print "." + "0.666".split(".")[-1] ..666 print "." + "123.666".split(".")[-1] ..666

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