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Home/ Questions/Q 8604171
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Editorial Team
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Editorial Team
Asked: June 12, 20262026-06-12T02:30:24+00:00 2026-06-12T02:30:24+00:00

Using data organised as dtl <- replicate(10,data.table(id=sample(letters,10),val=sample(10)), simplify=F) lapply(dtl, function(x){setkey(x,’id’)}) I need to extract

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Using data organised as

dtl <- replicate(10,data.table(id=sample(letters,10),val=sample(10)), simplify=F)
lapply(dtl, function(x){setkey(x,'id')})

I need to extract a list of datatables that contain the rows in dtl[[n+1]]] with id not present in dtl[[n]]. I assume it would be something like

dtl2 <- list(setdiff(dtl[[1]][['id']],dtl[[2]][['id']]),setdiff(dtl[[2]][['id']],dtl[[3]][['id']]...)

Please notice that, while the setdiff should only take the id column into account, I expect the result to contain all columns from each datatable.

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  1. Editorial Team
    Editorial Team
    2026-06-12T02:30:26+00:00Added an answer on June 12, 2026 at 2:30 am

    I think this will do it for you:

    mapply(setdiff, head(dtl, -1), tail(dtl, -1), SIMPLIFY = FALSE)
    

    Edit: with your new expected output, I would still use mapply as above, but with one of the following two changes:

    1. replace setdiff with function(x,y)setdiff(x$id, y$id)
    2. replace dtl with ids <- lapply(dtl, "[", "id")

    Edit2:: you’ve changed your expected output again by adding a plain English description that does not match the code you had provided… I think you are now looking for this:

    mapply(function(x,y)y[setdiff(y$id, x$id), ],
           head(dtl, -1), tail(dtl, -1), SIMPLIFY = FALSE)
    
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