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 09:38:57

I have changed the method to this:

list<time_period> splitTimePeriod(time_period timePeriod, int n)
{
time_duration equalPeriod = (timePeriod.end() - timePeriod.begin()) / 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++)
{

// tiempo de comienzo del siguiente período de tiempo
ptime tpNextBegin = tpBegin + equalPeriod;

// 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;
}

############################################################
I didn't know I can make these operations to time_duration:
time_duration equalPeriod = (timePeriod.end() - timePeriod.begin()) / n;

Now it works faster.

On Thu, Oct 2, 2014 at 10:21 AM, Pablo Madoery <madoerypablo_at_[hidden]>
wrote:

> 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
>>
>>
>> _______________________________________________
>> Boost-users mailing list
>> Boost-users_at_[hidden]
>> http://lists.boost.org/mailman/listinfo.cgi/boost-users
>>
>
>