data.table column order when using lapply and get

问题

can someone help me understand why the two versions of the lapply operations below with and without using get() don't produce the same result? When using get() the result columns get mixed up.

dt <- data.table(v1 = c(1,2), v2 = c(3,4), type = c('A', 'B'))

   v1 v2 type
1:  1  3    A
2:  2  4    B

col_in <- c('v2', 'v1')
col_out <- paste0(col_in, '.new')

accessing 'type' the hard-coded way

dt[, (col_out) := lapply(.SD, function(x){x * min(x[type == 'A'])}), .SDcols = col_in]

produces the expected result:

   v1 v2 type v2.new v1.new
1:  1  3    A      9      1
2:  2  4    B     12      2

however, when accessing 'type' via get()

dt[, (col_out) := lapply(.SD, function(x){x * min(x[get('type') == 'A'])}), .SDcols = col_in]

the expected values for v1.new are in v2.new and vice versa:

   v1 v2 type v2.new v1.new
1:  1  3    A      1      9
2:  2  4    B      2     12

Note: This a minimal toy example that I distilled down from a more complex operation that I'm trying to implement. The name of the 'type' variable is given as an input parameter.

回答1:

Interesting! Thanks for sharing! It seems that the use of get requires some internal sorting (bug?).

Two ways to avoid this:

Move the type == 'A' part outside the dt[,lapply(...)]

referenceRows <- which(dt[,type == 'A'])
referenceRows <- which(dt[,get('type') == 'A'])
dt[, lapply(.SD, function(x){x * min(x[referenceRows])}), .SDcols = col_in]

   v1 v2 type v2.new v1.new
1:  1  3    A      9      1
2:  2  4    B     12      2

First create the new columns and then use setnames to make sure that the new columns are assigned the proper columns names. Finally bind the two parts together with cbind:

dtNew <- dt[, lapply(.SD, function(x){x * min(x[type == 'A'])}), .SDcols = col_in]
setnames(dtNew, col_in, col_out)
cbind(dt, dtNew)


   v1 v2 type v2.new v1.new
1:  1  3    A      9      1
2:  2  4    B     12      2

Same result (although differently sorted):

    dtNew <- dt[, lapply(.SD, function(x){x * min(x[get('type') == 'A'])}), .SDcols = col_in]
    setnames(dtNew, col_in, col_out)
    cbind(dt, dtNew)


       v1 v2 type v1.new v2.new
    1:  1  3    A      1      9
    2:  2  4    B      2     12

回答2:

Another way is to use cool R feature called computing on the language (not related to data.table) instead of get and produce required j argument as language object using substitute function.
This will work also when grouping.

library(data.table)
dt <- data.table(v1 = c(1,2), v2 = c(3,4), type = c('A', 'B'))
col_in <- c('v2', 'v1')
col_out <- paste0(col_in, '.new')

col_where <- 'type'
qj <- substitute(.col_out := lapply(.SD, function(x){x * min(x[.col_where == 'A'])}),
                 list(.col_out=col_out, .col_where=as.name(col_where)))
print(qj)
#`:=`(c("v2.new", "v1.new"), lapply(.SD, function(x) {
#    x * min(x[type == "A"])
#}))

dt[, eval(qj), .SDcols = col_in][]
#      v1    v2   type v2.new v1.new
#   <num> <num> <char>  <num>  <num>
#1:     1     3      A      9      1
#2:     2     4      B     12      2

More about this nice feature in R language definition: Computing-on-the-language chapter.

来源：https://stackoverflow.com/questions/50878299/data-table-column-order-when-using-lapply-and-get

标签

data.table

lapply