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

Subject: Re: [Boost-users] [Boost] [DateTime] dividing a time_period in Ntime_periods
From: Igor Mironchik (igor.mironchik_at_[hidden])
Date: 2014-10-02 10:02:25

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_at_[hidden] 
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_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

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