Re: [Boost-users] [Boost] [DateTime] dividing a time_period in N time_periods

Subject: Re: [Boost-users] [Boost] [DateTime] dividing a time_period in N time_periods
From: Pablo Madoery (madoerypablo_at_[hidden])
Date: 2014-10-02 09:21:11

I think the time_duration s will be equal, but the time_periods will be
something like this:
2014-Jan-01 00:00:00 -> 2014-Jan-01 00:00:02.999999
2014-Jan-01 00:00:03 -> 2014-Jan-01 00:00:05.999999
2014-Jan-01 00:00:06 -> 2014-Jan-01 00:00:08.999999
2014-Jan-01 00:00:09 -> 2014-Jan-01 00:00:11.999999
2014-Jan-01 00:00:12 -> 2014-Jan-01 00:00:14.999999
2014-Jan-01 00:00:15 -> 2014-Jan-01 00:00:17.999999
2014-Jan-01 00:00:18 -> 2014-Jan-01 00:00:21

when calling the method like this

ptime p1 = dateTime::toPtime("2014-1-1 00:00:00");
ptime p2 = dateTime::toPtime("2014-1-1 00:00:21");

time_period tp(p1,p2);

list<time_period> l = dateTime::splitTimePeriod(tp, 7);
list<time_period>::iterator i = l.begin();
for(; i!= l.end(); ++i)
{
cout<<i->begin()<<" -> "<<i->end()<<endl;
}

On Thu, Oct 2, 2014 at 10:13 AM, Leon Mlakar <leon_at_[hidden]> wrote:

>  On 02/10/14 15:09, Pablo Madoery wrote:
>
> Hi, I want to make a method which divides a time_period in N time periods,
> something like this:
>
>  list<time_period> splitTimePeriod(time_period timePeriod, int n);
>
>  I actually have made it and although it seems to work it results very
> inefficient regarding the time it took to divide a large time_period.
>
>  The method I did is this:
>
>  list<time_period> splitTimePeriod(time_period timePeriod, int n)
> {
>  double secsTime = dateTime::diffSeconds(timePeriod.end(),
> timePeriod.begin());
>
>  // tiempo en segundos de cada período de tiempo
>  double ti = secsTime / n;
>
>  list<time_period> timePeriods;
>
>  // tiempo de comienzo del primer período de tiempo
>  ptime tpBegin = timePeriod.begin();
>
>  // tiempo de fin del último período de tiempo
>  ptime tpLastEnd = timePeriod.end();
>
>  for (int i = 0; i < n; i++)
>  {
>  int tUnits = ti * time_duration::ticks_per_second();
>
>  // tiempo de comienzo del siguiente período de tiempo
>  ptime tpNextBegin = dateTime::increment(tpBegin, tUnits);
>
>  // tiempo de fin del período de tiempo
>  ptime tpEnd;
>
>  if (i == n - 1)
>  {
>  tpEnd = tpLastEnd;
>  }
>  else
>  {
>  tpEnd = dateTime::decrement(tpNextBegin, 1);
>
>  // guarda para evitar que por cuestiones
>  // de redondeo, el tiempo final de un período
>  // sea mayor que el pasado inicialmente
>  // como argumento de entrada
>  if (tpEnd > tpLastEnd)
>  {
>  tpEnd = tpLastEnd;
>  }
>  }
>
>  time_period p(tpBegin, tpEnd);
>
>  // se inserta período de tiempo en la lista
>  timePeriods.push_back(p);
>
>  // se actualiza el comienzo del siguiente período de tiempo
>  tpBegin = tpNextBegin;
>  }
>
>  return timePeriods;
> }
>
>  Other methods I use are these:
>
>  double diffSeconds(const ptime &dateA, const ptime &dateB)
> {
>  time_duration diff = dateA - dateB;
>  return diff.total_seconds();
> }
>
>  ptime increment(const ptime &date, int n)
> {
>  time_iterator it(date, dateTime::unit);
>  for (int i = 0; i < n; i++)
>  {
>  ++it;
>  }
>  return *it;
> }
>
>  ptime decrement(const ptime &date, int n)
> {
>  time_iterator it(date, dateTime::unit);
>  for (int i = 0; i < n; i++)
>  {
>  --it;
>  }
>  return *it;
> }
>
>  const time_duration unit = microseconds(1);
>
>  #################################################
> The question is:  is there a better way to accomplish this?
>
>
> If you are splitting a time period into N periods of equal duration,
> wouldn't it be enough to return just one time period? The other n-1 are
> going to be equal, except perhaps for the last.
>
> Leon
>
>
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