Dear R-list users,
I have the following data frame, called MteBove:

posix_date    snow    temp
...
2010-01-20 23:30:00 45 NA
2010-01-02 00:30:00 10-2.7
2010-01-20 03:00:00 45 NA
2010-01-20 03:30:00 44 NA
2010-01-20 04:00:00 44 NA
2010-01-20 04:30:00 44 NA
2010-01-20 05:00:00 44 NA
2010-01-20 05:30:00 45 NA
2010-01-20 06:00:00 44 NA
2010-01-20 06:30:00 45 NA
...

> sapply(MteBove, class)
gives
$posix_date
[1] "POSIXct" "POSIXt"

$snow
[1] "numeric"

$temp
[1] "numeric"

There are semi-hourly data (i.e. 48 data each day) from the first of November 
to the first of May of each Winter season, and I need the daily mean for snow 
and temp.
First I created a subset for each Winter season
I tried to create a new column with only the day, like

> MteBove$day <- trunc(MteBove$posix_date, "days")

and then

> list_days <- unique(MteBove$day)
> means <- rapply(list_days, function(x) mean(MteBove[MteBove$day == x, ], 
> na.rm=T))

with the following result:
[1] "means IS"
  sec   min  hour  mday   mon  year  wday  yday isdst
   NA    NA    NA    NA    NA    NA    NA    NA    NA

I tried some changes always in this direction, with no success.
Is there an efficient way to do that?
Could somebody give me an hint about it?

Thank you for your attention and your help
Stefano Sofia


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