need to divide a date

Steve Holden steve at holdenweb.com
Thu Sep 22 16:24:05 CEST 2005


nephish at xit.net wrote:
> Hey there,
> i am doing a plotting application.
> i am using mxRelativeDateTimeDiff to get how much time is between
> date x and date y
> 
> now what i need to do is divide that time by 20 to get 20 even time
> slots
> for plotting on a graph.
> 
> For example, if the difference between them is 20 hours, i need 20
> plots, each an hour apart. if its 40 minutes, i need 20 plots that are
> 2 minutes apart.
> 
> what would be a way i could pull this off?
> 
> thanks
> 
C:\Steve\Projects\Python>python
Python 2.4.1 (#65, Mar 30 2005, 09:13:57) [MSC v.1310 32 bit (Intel)] on 
win32
Type "help", "copyright", "credits" or "license" for more information.
 >>> import mx.DateTime as dt
 >>> d1 = dt.DateTime(2005, 1, 1)
 >>> d2 = dt.DateTime(2005, 1, 21)
 >>> rdt = dt.RelativeDateTimeDiff(d2, d1)
 >>> rdt
<RelativeDateTime instance for 'YYYY-MM-(+20) HH:MM:SS' at 0x9f1300>
 >>> rdt/20
<RelativeDateTime instance for 'YYYY-MM-(+01) HH:MM:SS' at 0x9d5e90>
 >>> for i in range(20):
...    print dt.DateTime(2000) + i*(rdt/20)
...
C:\Python24\lib\site-packages\mx\DateTime\DateTime.py:585: 
DeprecationWarning: i
nteger argument expected, got float
   return DateTime(year, month, 1) + \
2000-01-01 00:00:00.00
2000-01-02 00:00:00.00
2000-01-03 00:00:00.00
2000-01-04 00:00:00.00
2000-01-05 00:00:00.00
2000-01-06 00:00:00.00
2000-01-07 00:00:00.00
2000-01-08 00:00:00.00
2000-01-09 00:00:00.00
2000-01-10 00:00:00.00
2000-01-11 00:00:00.00
2000-01-12 00:00:00.00
2000-01-13 00:00:00.00
2000-01-14 00:00:00.00
2000-01-15 00:00:00.00
2000-01-16 00:00:00.00
2000-01-17 00:00:00.00
2000-01-18 00:00:00.00
2000-01-19 00:00:00.00
2000-01-20 00:00:00.00
 >>>

Does this help?

regards
  Steve
-- 
Steve Holden       +44 150 684 7255  +1 800 494 3119
Holden Web LLC                     www.holdenweb.com
PyCon TX 2006                          www.pycon.org




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