monthly means with apply for multidimensional arrays - r

Question

I want to compute the mean over the 3-D of a multidimensional array. As this dimension is supposed to be the time, I wanted to computed monthly means. For that, I tried to use apply, but I am not sure where the problem is. Let's say my data is as the following:
#Creating a sample
m <-array(1:12, dim=c(20,4,36))
#number of months
months <- seq(1:12)
#Compute the mean over each month (dimension of the result should be [20,4,12]
monmean <- apply(m,1:2,function(x) for(i in 1:12) mean(x[,,months==i],na.rm=TRUE))
Any idea??
Thanks in advance

Answer 1

I think I understand what you're after. This is actually slightly more complex than it may seem, because months are not regular periods of time; they vary in number of days, and February varies between years due to leap years. Thus a simple regular logical or numeric index vector will not be sufficient to calculate this result precisely. You need to take into account the exact dates that are covered by the z-dimension of your array.
Solution 1
What you can do is separately compute a date vector that identifies the dates that correspond to each z-index of your array. Within the apply() call for each z-line, you can then call strftime() to extract the months for each such date, and group by that month value using tapply() to take monthly mean()s. Here's how it could be done:
set.seed(1);
R <- 48;
C <- 39;
Z <- 3653;
N <- R*C*Z;
a1 <- array(rnorm(N,10,2),c(R,C,Z));
dates <- seq(as.Date('2000-01-01'),as.Date('2009-12-31'),1);
a2 <- aperm(apply(a1,1:2,function(x) tapply(x,strftime(dates,'%m'),mean)),c(2,3,1));
Here's a demo showing a few specific proofs of correctness:
for (r in sample(1:nrow(a2),2)) for (c in sample(1:ncol(a2),2)) for (m in sample(1:dim(a2)[3],2)) cat(sprintf('[%02d,%02d,%3s] %f %f\n',r,c,month.abb[m],mean(a1[r,c,strftime(dates,'%m')==sprintf('%02d',m)]),a2[r,c,m]));
## [14,05,Aug] 10.030313 10.030313
## [14,05,Apr] 10.200982 10.200982
## [14,25,Jan] 9.957879 9.957879
## [14,25,Apr] 10.185447 10.185447
## [26,34,Oct] 10.056931 10.056931
## [26,34,Nov] 9.876327 9.876327
## [26,17,Apr] 10.005423 10.005423
## [26,17,Sep] 10.009785 10.009785
Notes
I randomly chose a date range of 2000-01-01 to 2009-12-31 because it covers a 10 year period during which (due to leap years) there were exactly 3653 days, but obviously you should be sure to use whatever dates are actually covered by your real data.
As you can see, you were on the right track by calling apply() with 1:2 as the margins, because that allows you to operate independently on each z-line, such that you can group that z-line by month and compute the mean for each month along that z-line.
Unfortunately, apply() has an annoying habit of returning the result in a different transposition than people generally expect. For two-dimensional usages, this is normally solved with a simple call to t(), but since we're working in three dimensions here, we need to call aperm() to fix the dimension order.
Since the dates I chose begin with January and advance through the months in calendar order, the means in the result will end up being ordered by calendar month. IOW, z-indexes 1:12 in a2 correspond to months Jan-Dec. If your dates do not begin with January, then this solution should still work, but you'll have to be careful about the correspondence between z-indexes and months in the result. For example, my "proof of correctness" code assumed that indexes 1:12 corresponded to months Jan-Dec, but that wouldn't be correct if the months occurred in a different order in the input array.
Solution 2
While writing this answer I actually thought of a slightly different, and one could argue slightly better, solution. You can call tapply() just once and group by rows, then columns, and finally months. Unfortunately, tapply() doesn't seem to be designed to naturally cycle its group vectors to cover the input vector, so we have to cycle them ourselves using carefully crafted calls to rep() (using the each and times arguments carefully--and I suppose tapply() actually wouldn't even know how to do this properly for our input data), but other than that, it's fairly straightforward:
a3 <- tapply(a1,list(rep(1:R,C*Z),rep(1:C,each=R,times=Z),rep(strftime(dates,'%m'),each=R*C)),mean);
Here's a proof that the result is identical to my first method (dimnames() have to be fixed first to get the identical() call to work, but that's trivial):
dimnames(a3) <- dimnames(a2);
identical(a3,a2);
## [1] TRUE
Performance
Here's some basic performance testing using system.time() to give an idea of the superiority of the second solution:
first <- function() a2 <- aperm(apply(a1,1:2,function(x) tapply(x,strftime(dates,'%m'),mean)),c(2,3,1));
second <- function() a3 <- tapply(a1,list(rep(1:R,C*Z),rep(1:C,each=R,times=Z),rep(strftime(dates,'%m'),each=R*C)),mean);
system.time({ first() });
## user system elapsed
## 3.672 0.015 3.719
system.time({ first() });
## user system elapsed
## 3.672 0.016 3.720
system.time({ second() });
## user system elapsed
## 1.797 0.344 2.135
system.time({ second() });
## user system elapsed
## 1.719 0.391 2.124