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R data.table by group replace first row of all missing column

Tags:

r

data.table

I have a data.table and I am trying to do something akin to data[ !is.na(variable) ]. However, for groups that are entirely missing, I'd like to just keep the first row of that group. So, I am trying to subset using by. I have done some research online and have a solution, but I think it is inefficient.

I've provided an example below showing what I am hoping to achieve, and I wonder if this can be done without creating the two extra columns.

d_sample = data.table( ID = c(1, 1, 2, 2, 3, 3), 
                   Time = c(10, 15, 100, 110, 200, 220), 
                   Event = c(NA, NA, NA, 1, 1, NA))

d_sample[ !is.na(Event), isValidOutcomeRow := T, by = ID]
d_sample[ , isValidOutcomePatient := any(isValidOutcomeRow), by = ID]
d_sample[ is.na(isValidOutcomePatient), isValidOutcomeRow := c(T, rep(NA, .N - 1)), by = ID]
d_sample[ isValidOutcomeRow == T ]

EDIT: Here are some speed comparisons with thelatemail and Frank's solutions with a larger dataset with 60K rows.

d_sample = data.table( ID = sort(rep(seq(1,30000), 2)), 
                   Time = rep(c(10, 15, 100, 110, 200, 220), 10000), 
                   Event = rep(c(NA, NA, NA, 1, 1, NA), 10000) )

thelatemail's solution gets a runtime of 20.65 on my computer.

system.time(d_sample[, if(all(is.na(Event))) .SD[1] else .SD[!is.na(Event)][1], by=ID])

Frank's first solution gets a runtime of 0

system.time( unique( d_sample[order(is.na(Event))], by="ID" ) )

Frank's second solution gets a runtime of 0.05

system.time( d_sample[order(is.na(Event)), .SD[1L], by=ID] )
like image 835
vryb Avatar asked Oct 17 '16 23:10

vryb


1 Answers

This seems to work:

unique( d_sample[order(is.na(Event))], by="ID" )

   ID Time Event
1:  2  110     1
2:  3  200     1
3:  1   10    NA

Alternately, d_sample[order(is.na(Event)), .SD[1L], by=ID].


Extending the OP's example, I also find similar timings for the two approaches:

n = 12e4 # must be a multiple of 6
set.seed(1)
d_sample = data.table( ID = sort(rep(seq(1,n/2), 2)), 
                   Time = rep(c(10, 15, 100, 110, 200, 220), n/6), 
                   Event = rep(c(NA, NA, NA, 1, 1, NA), n/6) )

system.time(rf <- unique( d_sample[order(is.na(Event))], by="ID" ))
# 1.17
system.time(rf2 <- d_sample[order(is.na(Event)), .SD[1L], by=ID] )   
# 1.24
system.time(rt <- d_sample[, if(all(is.na(Event))) .SD[1] else .SD[!is.na(Event)], by=ID])    
# 10.42
system.time(rt2 <- 
    d_sample[ d_sample[, { w = which(is.na(Event)); .I[ if (length(w) == .N) 1L else -w ] }, by=ID]$V1 ] 
)
# .13

# verify
identical(rf,rf2) # TRUE
identical(rf,rt) # FALSE
fsetequal(rf,rt) # TRUE
identical(rt,rt2) # TRUE

The variation on @thelatemail's solution rt2 is the fastest by a wide margin.

like image 94
Frank Avatar answered Sep 16 '22 14:09

Frank