I wanted the round up the number (5.0 = 5.0, not 6.0.). The ceil funciotn is
the right one for me.
Thanks to all.
>> Grant Edwards wrote:
>>> On 2005-08-30, Devan L <[EMAIL PROTECTED]> wrote:
>>> >
>>> > RoundToInt(2.0) will give you 3.
>>>
>>> That's what the OP said he wanted. The next bigger
On 2005-08-30, Devan L <[EMAIL PROTECTED]> wrote:
> Grant Edwards wrote:
>> On 2005-08-30, Devan L <[EMAIL PROTECTED]> wrote:
>> >
>> > RoundToInt(2.0) will give you 3.
>>
>> That's what the OP said he wanted. The next bigger integer
>> after 2.0 is 3.
>
> It's not really clear whether he wanted i
Grant Edwards wrote:
> On 2005-08-30, Devan L <[EMAIL PROTECTED]> wrote:
> >
> > RoundToInt(2.0) will give you 3.
>
> That's what the OP said he wanted. The next bigger integer
> after 2.0 is 3.
>
> --
> Grant Edwards grante Yow! I'd like TRAINED
>
On 2005-08-30, Devan L <[EMAIL PROTECTED]> wrote:
>>> Hi all. I'd need to aproximate a given float number into the
>>> next (int) bigger one. Because of my bad english I try to
>>> explain it with some example:
>>>
>>> 5.7 --> 6
>>> 52.987 --> 53
>>> 3.34 --> 4
>>> 2.1 --> 3
>>
>> The standard way
Thomas Bartkus wrote:
> On Sun, 28 Aug 2005 23:11:09 +0200, billiejoex wrote:
>
> > Hi all. I'd need to aproximate a given float number into the next (int)
> > bigger one. Because of my bad english I try to explain it with some example:
> >
> > 5.7 --> 6
> > 52.987 --> 53
> > 3.34 --> 4
> > 2.1 -->
On Sun, 28 Aug 2005 23:11:09 +0200, billiejoex wrote:
> Hi all. I'd need to aproximate a given float number into the next (int)
> bigger one. Because of my bad english I try to explain it with some example:
>
> 5.7 --> 6
> 52.987 --> 53
> 3.34 --> 4
> 2.1 --> 3
>
The standard way to do this is
Mikael Olofsson wrote:
> Michael Sparks wrote:
>
>> def approx(x):
>> return int(x+1.0)
>
> I doubt this is what the OP is looking for.
...
> Others have pointed to math.ceil, which is most likely what the OP wants.
I agree that's "likely" but, as Michael pointed out in the text you
removed
Michael Sparks wrote:
> def approx(x):
> return int(x+1.0)
I doubt this is what the OP is looking for.
>>> approx(3.2)
4
>>> approx(3.0)
4
Others have pointed to math.ceil, which is most likely what the OP wants.
/Mikael Olofsson
Universitetslektor (Senior Lecturer [BrE], Associate Profes
billiejoex wrote:
> Hi all. I'd need to aproximate a given float number into the next (int)
> bigger one. Because of my bad english I try to explain it with some
> example:
>
> 5.7 --> 6
> 52.987 --> 53
> 3.34 --> 4
> 2.1 --> 3
What about 2.0? By your spec that should be rounded to 3 - is that w
Thank you. :-)
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http://mail.python.org/mailman/listinfo/python-list
billiejoex wrote:
> Hi all. I'd need to aproximate a given float number into the next (int)
> bigger one. Because of my bad english I try to explain it with some example:
>
> 5.7 --> 6
> 52.987 --> 53
> 3.34 --> 4
> 2.1 --> 3
>
Have a look at math.ceil
>>> import math
>>> math.ceil(5.7)
6.0
billiejoex wrote:
> Hi all. I'd need to aproximate a given float number into the next (int)
> bigger one. Because of my bad english I try to explain it with some example:
>
> 5.7 --> 6
> 52.987 --> 53
> 3.34 --> 4
> 2.1 --> 3
>
> Regards
>
>
math.ceil returns what you need but as a float, th
Hi all. I'd need to aproximate a given float number into the next (int)
bigger one. Because of my bad english I try to explain it with some example:
5.7 --> 6
52.987 --> 53
3.34 --> 4
2.1 --> 3
Regards
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