data.table and stratified means

Question

I've got some code that generate stratified weighted means and I'm certain this worked a few months ago. But, but I'm not sure what the current problem is. (I apologize - this must be very basic stuff):

dp=
structure(list(seqn = c(1L, 2L, 3L, 4L, 6L, 7L, 8L, 9L, 10L, 
11L, 12L, 13L, 3L, 4L, 9L, 10L, 11L, 14L, 8L, 11L, 12L, 10L, 
5L, 13L, 2L, 14L, 3L, 9L, 6L, 7L), sex = c(2L, 1L, 2L, 2L, 1L, 
2L, 2L, 1L, 2L, 2L, 2L, 1L, 1L, 2L, 2L, 1L, 1L, 2L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), bmi = c(22.8935608711259, 
27.0944623781918, 40.4637162938634, 23.7649712675423, 15.3193372705538, 
31.1280302540991, 21.4866354393239, 20.3200254374398, 32.331092513536, 
25.3679771839413, 33.9400508162971, 14.7048592172926, 25.5243757788688, 
23.4331882363495, 27.6428134168995, 29.3923629426172, 24.9547209666314, 
17.0522203606383, 15.51, 22, 30.62, 30.94, 29.1, 25.57, 24.9, 
27.33, 17.63, 18.48, 22.56, 29.39), tc = c(273L, 181L, 150L, 
201L, 142L, 165L, 235L, 219L, 298L, 222L, 143L, 134L, 268L, 160L, 
236L, 225L, 260L, 140L, 162L, 132L, 156L, 140L, 279L, 314L, 215L, 
174L, 129L, 148L, 153L, 245L), swt = c(1645, 3318, 2280, 1574, 
4062, 1627, 14604, 24675, 975, 975, 2697, 1559, 1737.58, 1730.23, 
19521.36, 28080.57, 1248.43, 13745.77, 5251.76464426326, 6497.194885522, 
15915.7023420765, 3740.96809540218, 16574.177622509, 307.32513798849, 
4720.89748295751, 3247.78896499604, 7698.70949077031, 1262.6450411464, 
6609.43340735515, 4254.23723479882)), .Names = c("seqn", "sex", 
"bmi", "tc", "swt"), row.names = c(20560L, 20561L, 20562L, 20563L, 
20565L, 20566L, 20567L, 20568L, 20569L, 20570L, 20571L, 20572L, 
61335L, 61336L, 61338L, 61339L, 61340L, 61341L, 95465L, 96890L, 
104613L, 105988L, 107581L, 112267L, 113403L, 114292L, 119979L, 
120271L, 125939L, 135699L), class = "data.frame")

dt=data.table(dp, key='sex')

sapply(df,function(x)weighted.mean(x,df$swt))  #this works to weighted mean
dt[,lapply(.SD, mean, na.rm=T), .SDcols=c('bmi','tc','swt')]  
     #this also works for overall unweighted mean

dt[,lapply(.SD, function(x)weighted.mean(x,swt, na.rm=TRUE)), by=key(dt), .SDcols=c('bmi','tc','swt')] 

but this gives the error: Error in weighted.mean.default(x, swt, na.rm = TRUE) : object 'swt' not found

sessionInfo()
R version 2.15.2 (2012-10-26)
Platform: i386-w64-mingw32/i386 (32-bit)

locale:
[1] LC_COLLATE=English_United States.1252  LC_CTYPE=English_United States.1252   
[3] LC_MONETARY=English_United States.1252 LC_NUMERIC=C                          
[5] LC_TIME=English_United States.1252    

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] data.table_1.8.6

loaded via a namespace (and not attached):
[1] tools_2.15.2

Answer 1

UPDATE (from Arun): This is now fixed in v1.8.11. From NEWS:

o DT[, lapply(.SD, function(), by=] did not see columns of DT when optimisation is "on". This is now fixed, #2381. Tests added and tested successfully. Thanks to David F for reporting on SO: data.table and stratified means

---

This is indeed a bug introduced somewhere between 1.8.2 and 1.8.6.

dt[,lapply(.SD, function(x) weighted.mean(x,swt, na.rm=TRUE)), by=key(dt),
    .SDcols=c('bmi','tc','swt')] 
Error in weighted.mean.default(x, swt, na.rm = TRUE) : 
    object 'swt' not found

To work around this in the meantime, either turn off optimization :

options(datatable.optimize=FALSE)
dt[,lapply(.SD, function(x)weighted.mean(x,swt, na.rm=TRUE)), by=key(dt),    
    .SDcols=c('bmi','tc','swt')]
   sex      bmi       tc      swt
1:   1 25.64376 206.0115 17171.20
2:   2 23.73566 193.8727 11467.47

or, don't wrap with function() :

options(datatable.optimize=TRUE)
dt[,lapply(.SD, weighted.mean, swt, na.rm=TRUE), by=key(dt),    
    .SDcols=c('bmi','tc','swt')] 
   sex      bmi       tc      swt
1:   1 25.64376 206.0115 17171.20
2:   2 23.73566 193.8727 11467.47

We are making more use of optimization now, but this case slipped through the test suite: tests 825.1, 825.2 and 825.3 didn't cover an argument to a function being another column, within an anonymous function(). It would be a problem where the function isn't already given; i.e., unlike this case, where the function() can just be omitted since weighted.mean is already given and can be applied as-is.

You can see how optimization modifies j by setting verbose=TRUE (either per query or with the global option). In this case nothing would have been revealed as wrong by that verbose output, but just mentioning it as an aside.

Now filed as #2381: Optimization of lapply(.SD, function() ...) no longer sees columns inside .... Will fix and add tests so this can't regress again.

Thanks!

Answer 2

I suggest to keep it simple:

dt[,list(bmi_m=weighted.mean(bmi,swt,na.rm=TRUE),
         tc_m=weighted.mean(tc,swt,na.rm=TRUE),
         swt_m=weighted.mean(swt,swt,na.rm=TRUE)),by=key(dt)]

I think this is also reasonably fast.