delta time = time stop - time start

[Guilherme Salgado]

On Mon, 2004-01-26 at 04:08, engsol wrote:
[...]
> The routine I came up with is below, but seems a bit clunky.
> Is there a better way of doing this? I think it relies too much on integers rounding off in the proper direction, a la
> d_time_hr = d_time / 3600 below.
> 

You can use mx.DateTime
(http://www.lemburg.com/files/python/mxDateTime.html).
Then, the code will look like this:

from mx import DateTime
start = DateTime.now()
# do something ...
stop = DateTime.now()
delta = (stop - start).strftime("%H:%M:%S")

hope it helps.

> Also, this will have to transition to Linux, if that makes a difference.
> 
> START CODE:
> 
> import string
> 
> def deltatime(start, stop):
> 
>     t_start = (string.atoi(start[0:2]) * 3600) + (string.atoi(start[3:5]) * 60) + string.atoi(start[6:8])
>     t_stop  = (string.atoi(stop [0:2]) * 3600) + (string.atoi(stop [3:5]) * 60) + string.atoi(stop [6:8])
> 
>     if t_start < t_stop:
>         d_time = t_stop - t_start
>     else:
>         d_time = (86400 - t_start) + t_stop
> 
>     d_time_hr  =  d_time / 3600
>     d_time_min = (d_time - d_time_hr * 3600) / 60
>     d_time_sec = (d_time - d_time_hr * 3600) - (d_time_min * 60)
> 
>     return str(d_time_hr) + 'hr ' + str(d_time_min) + 'min ' + str(d_time_sec) + 'sec'
> 
> END CODE

[]'s
Salgado
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