Try this:

sapply(0:(length(DateTime3)-1),  function(i)as.character(strptime(start,
"%m/%d/%Y %H:%M") + i * interval))


On Wed, Oct 13, 2010 at 8:51 AM, dpender <d.pender@civil.gla.ac.uk> wrote:

>
> I am trying to convert an array from numeric values back to date and time
> format.  The code I have used is as follows;
>
> for (i in 0:(length(DateTime3)-1))      {
>        DateTime3[i] <- (strptime(start, "%m/%d/%Y %H:%M")+ i*interval)
>
> where start <- [1] "1/1/1981 00:00"
>
> However the created array (DateTime3) contains [1,] 347156400 347157600
> 347158800 347160000 347161200 347162400 347163600   NA
>
> Does anyone know how I can change DateTime 3 to the same format as start?
>
> Thanks,
>
> Doug
>
> --
> View this message in context:
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>
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>



-- 
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O

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