Reputation: 301
Firstly, let us generate data like this:
library(data.table)
data <- data.table(date = as.Date("2015-05-01")+0:299)
set.seed(123)
data[,":="(
a = round(30*cumprod(1+rnorm(300,0.001,0.05)),2),
b = rbinom(300,5000,0.8)
)]
Then I want to use my custom function to operate multiple columns multiple times without manually typing out .Such as my custom function is add <- function(x,n) (x+n)
I provide my for loops code as following:
add <- function(x,n) (x+n)
n <- 3
freture_old <- c("a","b")
for(i in 1:n ){
data[,(paste0(freture_old,"_",i)) := add(.SD,i),.SDcols =freture_old ]
}
Could you please tell me a lapply
version to instead of for loop?
Upvotes: 3
Views: 528
Reputation: 25225
An option without R "loop" (quoted since ultimately its a loop at certain level somewhere):
data[,
c(outer(freture_old, seq_len(n), paste, sep="_")) :=
as.data.table(matrix(outer(as.matrix(.SD), seq_len(n), add), .N)),
.SDcols=freture_old]
Or equivalently in base R:
setDF(data)
cbind(data, matrix(outer(as.matrix(data[, freture_old]), seq_len(n), add),
nrow(data)))
Upvotes: 1
Reputation: 451
If all you want is to use an lapply
loop instead of a for loop you really do not need to change much. For a data.table
object it is even easier since every iteration will change the data.table
without having to save a copy to the global environment. One thing I add just to suppress the output to the console is to wrap an invisible
around it.
lapply(1:n,function(i) data[,paste0(freture_old,"_",i):=lapply(.SD,add,i),.SDcols =freture_old])
Note that if you assign this lapply to an object you will get a list of data.tables
the size of the number of iterations or in this case 3. This will kill memory because you are really only interested in the final entry. Therefore just run the code without assigning it to a variable. Now if you do not assign it to anything you will get every iteration printed out to the console. So what I would suggest is to wrap an invisible
around it like this:
invisible(lapply(1:n,function(i) data[,paste0(freture_old,"_",i):=lapply(.SD,add,i),.SDcols =freture_old]))
Hope this helps and let me know if you need me to add anything else to this answer. Good luck!
Upvotes: 5