Functions with Arrays in R - arrays

Question

Let's say I have maximum temperature data for the last 20 years. My data frame has a column for month, day, year and MAX_C (temperature data). I want to calculate the mean (and standard deviation, and range) maximum temperature from June 31 of one year to July 1 of the preceding year (i.e. mean max daily temp from July 1, 1991 to June 31, 1992). Is there an efficient way to do this?
My approach, thus far, has been to create an array:
maxt.prev12<-tapply(maxt$MAX_C,INDEX=list(maxt$month,maxt$day,maxt$year),mean)
I put mean in as the function as tapply was not producing an array without a function after the INDEX, but mean is not actually calculating anything here. Then I was thinking about trying to take January through June from one the matrices (i.e. 1992), and July through December from the preceding matrix (i.e. 1991), and then computing the mean. I'm not entirely sure how to do that part, however, there must be a more efficient way of performing these calculations in R
EDIT
Here is a simple sample set of data
maxt
day month year MAX_C
1 1 1990 29
1 2 1990 28
1 3 1990 32
1 4 1990 26
1 5 1990 24
1 6 1990 32
1 7 1990 30
1 8 1990 28
1 9 1990 28
1 10 1990 24
1 11 1990 30
1 12 1990 30
1 1 1991 25
1 2 1991 26
1 3 1991 28
1 4 1991 25
1 5 1991 24
1 6 1991 32
1 7 1991 26
1 8 1991 32
1 9 1991 26
1 10 1991 26
1 11 1991 27
1 12 1991 26
1 1 1992 27
1 2 1992 25
1 3 1992 29
1 4 1992 32
1 5 1992 27
1 6 1992 27
1 7 1992 24
1 8 1992 25
1 9 1992 28
1 10 1992 26
1 11 1992 31
1 12 1992 27

Answer 1

I would create an "indicator year" column which was equal to the year if month in July-Dec but equal to year-1 when month in Jan-June.
EDITED month reference in light of the fact it was numeric rather than character:
> maxt$year2 <- maxt$year
> maxt[ maxt$month %in% 1:6, "year2"] <-
+ maxt[ maxt$month %in% 1:6, "year"] -1
> # month.name is a 12 element constant vector in all versions of R
> # check that it matches the spellings of your months
>
> mean_by_year <- tapply(maxt$MAX_C, maxt$year2, mean, na.rm=TRUE)
> mean_by_year
1989 1990 1991 1992
28.50000 27.50000 27.50000 26.83333
If you wanted to change the labels so they reflected the non-calendar year derivation:
> names(mean_by_year) <- paste(substr(names(mean_by_year),3,4),
+ as.character( as.numeric(substr(names(mean_by_year),3,4))+1),
sep="_")
> mean_by_year
89_90 90_91 91_92 92_93
28.50000 27.50000 27.50000 26.83333
Although I don't think it will be quite right at the millennial turn.