I am performing a subset of a large ffdf objects and I noticed that when I use subset.ff it is generating a large number of NAs. I tried an alternative way by using ffwhich and the result is much faster and no NAs are generated. Here it is my test:
library(ffbase)
# deals is the ffdf I would like to subset
unique(deals$COMMODITY)
ff (open) integer length=7 (7) levels: CASH CO2 COAL ELEC GAS GCERT OIL
[1] [2] [3] [4] [5] [6] [7]
CASH CO2 COAL ELEC GAS GCERT OIL
# Using subset.ff
started.at=proc.time()
deals0 <- subset.ff(deals,deals$COMMODITY %in% c("CASH","COAL","CO2","ELEC","GCERT"))
cat("Finished in",timetaken(started.at),"\n")
Finished in 12.640sec
# NAs are generated
unique(deals0$COMMODITY)
ff (open) integer length=8 (8) levels: CASH CO2 COAL ELEC GAS GCERT OIL <NA>
[1] [2] [3] [4] [5] [6] [7] [8]
CASH CO2 COAL ELEC GAS GCERT OIL NA
# Subset using ffwhich
started.at=proc.time()
idx <- ffwhich(deals,COMMODITY %in% c("CASH","COAL","CO2","ELEC","GCERT"))
deals1 <- deals[idx,]
cat("Finished in",timetaken(started.at),"\n")
Finished in 3.130sec
# No NAs are generated
unique(deals1$COMMODITY)
ff (open) integer length=7 (7) levels: CASH CO2 COAL ELEC GAS GCERT OIL
[1] [2] [3] [4] [5] [6] [7]
CASH CO2 COAL ELEC GAS GCERT OIL
Any idea why this is happening?
subset.ff is probably using [and your criterion but not including a !is.na(.) clause. The default for "[" is to return items that are either TRUE or NA for the criterion vector. The regular subset function adds a !is.na(.) clause, but maybe the authors of ffbase didn't get around to that.