使用 data.table 和 lapply 的变量中的列名

Question

I have data similar to this:

set.seed(1)
testing1 <- data.table(type=c("stock","stock","bond","bond"),a=rnorm(4),b=rnorm(4),c=rnorm(4),d=rnorm(4),e=rnorm(4))

    type          a          b          c           d           e
1: stock -0.6264538  0.3295078  0.5757814 -0.62124058 -0.01619026
2: stock  0.1836433 -0.8204684 -0.3053884 -2.21469989  0.94383621
3:  bond -0.8356286  0.4874291  1.5117812  1.12493092  0.82122120
4:  bond  1.5952808  0.7383247  0.3898432 -0.04493361  0.59390132

This returns exactly what I want:

result1 <- testing1[,c(list(type=type),lapply(.SD, `-`, a)), .SDcols = b:e]

    type          b          c           d          e
1: stock  0.9559616  1.2022352  0.00521323  0.6102635
2: stock -1.0041117 -0.4890317 -2.39834321  0.7601929
3:  bond  1.3230577  2.3474098  1.96055953  1.6568498
4:  bond -0.8569561 -1.2054376 -1.64021441 -1.0013795

The problem is that the column a is dynamically named. I would like to do something like this:

cn <- "a"
result2 <- testing1[,c(list(type=type), lapply(.SD, `-`, get(cn))), .SDcols = b:e]

but I get the error message: Error in FUN(X[[i]], ...) : non-numeric argument to binary operator .

Any ideas would be greatly appreciated. Thanks.

Answer 1

We can make use of [[ to extract the column from 'testing1'. Here .SD would not work as the column is not specified in .SDcols

testing1[,c(list(type=type),lapply(.SD, `-`, testing1[[cn]])), .SDcols = b:e]
#   type          b          c           d          e
#1: stock  0.9559616  1.2022352  0.00521323  0.6102635
#2: stock -1.0041117 -0.4890317 -2.39834321  0.7601929
#3:  bond  1.3230577  2.3474098  1.96055953  1.6568498
#4:  bond -0.8569561 -1.2054376 -1.64021441 -1.0013795

---

If we are using get , make sure the environment is the same as with lapply , the environment is from .SD which doesn't have the column 'a'. Instead use Map

testing1[,  Map(`-`, .SD, list(get(cn))), .SDcols = b:e]