Why rollapply is much slower with fill=NA and function return multiple values?

Question

If a function return 2 or more values, and using fill = NA , rollapply become much slower. Is there any ways to avoid it?

f1= function(v)c(mean(v)+ median(v))    #return vector of length 1
f2= function(v)c(mean(v), median(v))    #return vector of length 2


v = rnorm(1000)
microbenchmark(rollapplyr(v, 20, f1), rollapplyr(v,20, f1, fill=NA) )

#                             expr      min       lq     mean   median       uq      max neval
#            rollapplyr(v, 20, f1) 50.84485 53.68726 57.21892 54.63793 57.78519 75.88305   100
# rollapplyr(v, 20, f1, fill = NA) 52.11355 54.69866 59.73473 56.20600 63.10546 99.96493   100

microbenchmark(rollapplyr(v, 20, f2), rollapplyr(v,20, f2, fill=NA) )

#                             expr      min       lq     mean   median       uq      max neval
#            rollapplyr(v, 20, f2) 51.77687 52.29403 56.80307 53.44605 56.65524 105.6713   100
# rollapplyr(v, 20, f2, fill = NA) 69.93853 71.08953 76.48056 72.21896 80.58282 151.4455   100

Answer 1

The reason is to be found in the speed of using fill.na on different types of data, as happens internally in the rollapply() function. Your f1 returns a single vector, whereas f2 returns a matrix of two columns (well, both are zoo objects actually, but you catch my drift).

The speed decrease for inserting the NA is not proportionate to the mere doubling of the number of elements, as this shows:

library(zoo)
library(microbenchmark)

v <- zoo(rnorm(1000))
m <- zoo(matrix(rnorm(2000), ncol=2))
ix <- seq(1000)>50
microbenchmark(na.fill(v, NA, ix), na.fill(m, NA, ix))

# Unit: microseconds
#               expr      min        lq       mean    median         uq      max neval
# na.fill(v, NA, ix)  402.861   511.912   679.1114   659.597   754.8385  4716.46   100
# na.fill(m, NA, ix) 9746.643 10091.038 14281.5598 14057.304 17589.9670 22249.96   100