Aggregate 5 minute data to hourly sums with NA's - r

Question

My problem is as follows: I've got a time series with 5-Minute precipitation data like:
Datum mm
1 2004-04-08 00:05:00 NA
2 2004-04-08 00:10:00 NA
3 2004-04-08 00:15:00 NA
4 2004-04-08 00:20:00 NA
5 2004-04-08 00:25:00 NA
6 2004-04-08 00:30:00 NA
With this structure:
'data.frame': 1098144 obs. of 2 variables:
$ Datum: POSIXlt, format: "2004-04-08 00:05:00" "2004-04-08 00:10:00" "2004-04-08 00:15:00" "2004-04-08 00:20:00" ...
$ mm : num NA NA NA NA NA NA NA NA NA NA ...
As you can see, the time series begins with a lot of NA's, but there is measured precipitation further down, although riddled with single, less common NA's due to malfunction of the measuring station.
What I'm trying to achieve, is summing up the measured precipitation to hourly sums, not considering NA's.
This is what I tried so far:
sums <- aggregate(precip["mm"],
list(cut(precip$Datum, "1 hour")), sum)
Even though the timestamps are correctly aggregated to hours, all sums are 0 or NA. The sums are not even calculated if there is no NA at all.
additionally to be taken into account:
Hourly precipitation sums in meteorology always describe the cumulative sum until a certain hour: The amount of precipitation at 0:00 o'clock describes the sum from 23:00 the previous day until 0:00. So I always need to sum up the previous hour.
Reproducible Example
set.seed(1120)
s <- as.POSIXlt("2004-03-08 23:00:00")
r <- seq(s, s+1e4, "30 min")
precip <- data.frame(Datum=r, mm=sample(c(1:5,NA), 6, T))
Datum mm
2004-03-08 23:00:00 4
2004-03-08 23:30:00 1
2004-03-09 00:00:00 2
2004-03-09 00:30:00 4
2004-03-09 01:00:00 1
2004-03-09 01:30:00 4
With the above example, the result I am looking for is:
Datum mm
2004-03-09 00:00:00 5
2004-03-09 01:00:00 6
2004-03-09 02:00:00 5

Answer 1

Try adding na.rm=TRUE:
aggregate(precip['mm'], list(cut(precip$Datum, "1 hour")), sum, na.rm=TRUE)
# Group.1 mm
# 1 2004-04-08 00:00:00 26
# 2 2004-04-08 01:00:00 35
# 3 2004-04-08 02:00:00 25
Reproducible Example
set.seed(1120)
s <- as.POSIXlt("2004-04-08 00:05:00")
r <- seq(s, s+1e4, "5 min")
precip <- data.frame(Datum=r, mm=sample(c(1:5,NA), 34, T))
addendum
To your second question: If you would like measurements on the hour to be calculated with the lesser hour add right=TRUE:
aggregate(precip['mm'], list(cut(precip$Datum, "1 hour", right=TRUE)), sum, na.rm=TRUE)
Further Explanation
We will create another more detailed explanation to show how the solution works:
p <- c("2004-04-07 23:48:20", "2004-04-08 00:00:00", "2004-04-08 00:03:20")
ptime <- as.POSIXlt(p)
#[1] "2004-04-07 23:48:20 EDT" "2004-04-08 00:00:00 EDT" "2004-04-08 00:03:20 EDT"
We have three dates to separate into groups. If we use cut without any extra arguments, the second entry "2004-04-08 00:00:00 EDT" will be grouped with the third entry for hour "00:00":
cut(ptime, "1 hour")
#[1] 2004-04-07 23:00:00 2004-04-08 00:00:00 2004-04-08 00:00:00
But if we add the argument right=FALSE we can group it with the "23:00" hour:
cut(ptime, "1 hour", right=TRUE)
#[1] 2004-04-07 23:00:00 2004-04-07 23:00:00 2004-04-08 00:00:00
We can specify the behavior of edge cases.
edit
With your new data the original solution produces the desired output:
aggregate(precip['mm'], list(cut(precip$Datum, "1 hour")), sum, na.rm=TRUE)
Group.1 mm
1 2004-03-08 23:00:00 5
2 2004-03-09 00:00:00 6
3 2004-03-09 01:00:00 5

Answer 2

You can use dplyr to calculate sum like :
precip$hour <- strftime(precip$Datum,"%Y-%m-%d %H")
library(dplyr)
sum_hour <- precip %>% group_by(hour) %>% summarise(sum_hour = sum(mm,na.rm = T))