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: Leon Mlakar (leon_at_[hidden])
Date: 2014-10-02 09:13:29

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