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
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