Rounding time algo

[Marcus]

Hi, was wondering if anyone knows a good algo for rounding time to a
desired interval.

For instance, I have data set that is timestamped in second resolution
and I need to modify that data to be timestamped for an x minute
interval. So if time t = 93601, and x = 5 then t would be rounded up
to 94000... if x were 1, t would equal 93700, etc. All time is integer
based.

Any help would be really appreciated.

Marcus

[Mark P]

Re: Rounding time algo

Marcus wrote:

Hi, was wondering if anyone knows a good algo for rounding time to a
desired interval.
>
For instance, I have data set that is timestamped in second resolution
and I need to modify that data to be timestamped for an x minute
interval. So if time t = 93601, and x = 5 then t would be rounded up
to 94000... if x were 1, t would equal 93700, etc. All time is integer
based.
>
Any help would be really appreciated.
>
Marcus
>

I don't understand your math. If x = 5 and you want to round to 5
minute intervals, then a value t in seconds should be a multiple of 5 *
60 = 300 seconds. But 94000 is not divisible by 300. It looks like
you're assuming 100 seconds per minute based on your rounded values.

In any event, we can phrase this as a more general question: how do you
round a value n to a multiple of m? Since it looks like you want to
round up, you can do the following:

int remainder = n % m;
int rounded_n = n + (remainder ? (m - remainder) : 0);

[Marcus]

Re: Rounding time algo

Mark P wrote:

Marcus wrote:

Hi, was wondering if anyone knows a good algo for rounding time to a
desired interval.

For instance, I have data set that is timestamped in second resolution
and I need to modify that data to be timestamped for an x minute
interval. So if time t = 93601, and x = 5 then t would be rounded up
to 94000... if x were 1, t would equal 93700, etc. All time is integer
based.

Any help would be really appreciated.

Marcus

>
I don't understand your math. If x = 5 and you want to round to 5
minute intervals, then a value t in seconds should be a multiple of 5 *
60 = 300 seconds. But 94000 is not divisible by 300. It looks like
you're assuming 100 seconds per minute based on your rounded values.
>
In any event, we can phrase this as a more general question: how do you
round a value n to a multiple of m? Since it looks like you want to
round up, you can do the following:
>
int remainder = n % m;
int rounded_n = n + (remainder ? (m - remainder) : 0);

Hi Mark, thanks for the reply.

Reading back over my post I can see where the confusion is. Time t is
expressed as an integer, however it reflects an actual timestamp
containing hours minutes and seconds, instead of an integer expressed
in seconds. So the above 93601, can also be written 9:36:01.

Not sure if that would change your approach. What exactly are your
variables there, I'm not quite sure I understand? The other little
problem with straight division are the hour increments every 60 min...
that's why I was wondering if there might be a special approach to
this.

Marcus

[Mark P]

Re: Rounding time algo

Marcus wrote:

Mark P wrote:

>Marcus wrote:

>>Hi, was wondering if anyone knows a good algo for rounding time to a
>>desired interval.
>>>
>>For instance, I have data set that is timestamped in second resolution
>>and I need to modify that data to be timestamped for an x minute
>>interval. So if time t = 93601, and x = 5 then t would be rounded up
>>to 94000... if x were 1, t would equal 93700, etc. All time is integer
>>based.
>>>
>>Any help would be really appreciated.
>>>
>>Marcus
>>>

>I don't understand your math. If x = 5 and you want to round to 5
>minute intervals, then a value t in seconds should be a multiple of 5 *
>60 = 300 seconds. But 94000 is not divisible by 300. It looks like
>you're assuming 100 seconds per minute based on your rounded values.
>>
>In any event, we can phrase this as a more general question: how do you
>round a value n to a multiple of m? Since it looks like you want to
>round up, you can do the following:
>>
>int remainder = n % m;
>int rounded_n = n + (remainder ? (m - remainder) : 0);

>
Hi Mark, thanks for the reply.
>
Reading back over my post I can see where the confusion is. Time t is
expressed as an integer, however it reflects an actual timestamp
containing hours minutes and seconds, instead of an integer expressed
in seconds. So the above 93601, can also be written 9:36:01.
>
Not sure if that would change your approach. What exactly are your
variables there, I'm not quite sure I understand? The other little
problem with straight division are the hour increments every 60 min...
that's why I was wondering if there might be a special approach to
this.
>
Marcus
>

I explain what n and m are in the preceding text. remainder is a
temporary variable. rounded_n is, as you might guess, the rounded value
of n.

One straightforward way to work with times is convert everything to
seconds, perform the required arithmetic, and then convert everything
back. I'll let you figure out the details of how to do this but you'll
find the mod operator helpful for both directions of the conversion.

[Frederick Gotham]

Re: Rounding time algo

Marcus posted:

Hi, was wondering if anyone knows a good algo for rounding time to a
desired interval.

No, but if I exert my brain for approximately seven seconds, I'd probably
come up with one. You should try it.

For instance, I have data set that is timestamped in second resolution
and I need to modify that data to be timestamped for an x minute
interval. So if time t = 93601, and x = 5 then t would be rounded up
to 94000... if x were 1, t would equal 93700, etc. All time is integer
based.

You shouldn't have mentioned time at all. All you want to do is round an
integer up to the nearest specified integer.

Any help would be really appreciated.

template<class IntType>
inline
IntType RoundUp(IntType const x,IntType const factor)
{
return factor * (x/factor + !!(x%factor));
}

Or, if you would prefer a macro:

#define ROUND_UP_RAW(x, factor) (factor * (x/factor + !!(x%factor)))
#define ROUND_UP(x,fact or) ROUND_UP_RAW((x ),(factor))

--

Frederick Gotham

[Jim Langston]

Re: Rounding time algo

"Marcus" <mcdesigns@wall a.comwrote in message
news:1158213290 .304567.14730@m 73g2000cwd.goog legroups.com...

Hi, was wondering if anyone knows a good algo for rounding time to a
desired interval.
>
For instance, I have data set that is timestamped in second resolution
and I need to modify that data to be timestamped for an x minute
interval. So if time t = 93601, and x = 5 then t would be rounded up
to 94000... if x were 1, t would equal 93700, etc. All time is integer
based.
>
Any help would be really appreciated.
>
Marcus

Rounding an integer is fairly simple. If you want to drop 2 places, then
divide by 100 and mulitply by 100. Since integer division drops any
fraction/remainder you get what you want.

93601 / 100 == 936
936 * 100 = 93600

Now, we get into the complication of rounding. Say we wanted to loose 3
places of presion.

93601 / 1000 = 93
93 * 1000 = 93000

which is not what you want. The normal way to handle rounding off is to add
some muliple of 5. Since we want to round the 100's to the nearest
thousand, add 500.

( 93601 + 500 ) == 94101 / 1000 = 94
94 * 1000 == 94000

An actual function to do this is trivial.