And one more... If you use C++11 then better to use emplace_back() instead of push_back()... From: Igor Mironchik Sent: Thursday, October 02, 2014 5:02 PM To: boost-users@lists.boost.org Subject: Re: [Boost-users] [Boost] [DateTime] dividing a time_period in Ntime_periods Hi. Try this: list< time_period > splitTimePeriod( time_period timePeriod, int n ) { const 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 const ptime tpLastEnd = timePeriod.end(); for( int i = 0; i < n; ++i ) { // tiempo de fin del período de tiempo ptime tpEnd; if( i == n - 1 ) tpEnd = tpLastEnd; else { tpEnd = tpBegin + equalPeriod; // 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 = tpEnd; } return timePeriods;} It should be a little be more efficient. And I think that this is excess: // guarda para evitar que por cuestiones// de redondeo, el tiempo final de un período// sea mayor que el pasado inicialmente// como argumento de entradaif( tpEnd > tpLastEnd ) tpEnd = tpLastEnd; From: Pablo Madoery Sent: Thursday, October 02, 2014 4:38 PM To: boost-users@lists.boost.org Subject: Re: [Boost-users] [Boost] [DateTime] dividing a time_period in Ntime_periods 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@gmail.com> 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@digiverse.si> 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@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users -------------------------------------------------------------------------------- _______________________________________________ Boost-users mailing list Boost-users@lists.boost.org http://lists.boost.org/mailman/listinfo.cgi/boost-users