Calling user defined function from data.table object

Question

I am trying to call a user defined function in order to create a new column that depends on the value of the other columns of my data.table. In simple cases, I do not encounter any error, but when I am either using conditional statement or loops, it looks as if the user defined function receives the entire column as a parameter.

Learning from other cases reported on stack overflow (eg : R data.table user defined function), I understand that this problem can be overcome for if statements using the ifelse function. However, I can't find a solution for the loop statement.

Please, see below the code I want to run that returns the following error message : Error in seq.default(1, a, 1) : 'to' must be of length 1

test <-data.table(a=c(1,2))

f <- function(a) {
  out <- 0
  for (i in seq(1,a,1)){
    out <- out +1
  }
  return(out)
}

test[,b:=f(a)]

Obviously, f(x)=x but I chose this function for the sake of simplicity. Also, note that replacing seq(1,a,1) by 1:a throws the following warning message : In 1:a : numerical expression has 2 elements: only the first used.

Below is more detailed explanation of the desired behavior.

test <-data.table(a=c(1,2,3),b=c(4,5,6))
f <- function(a,b){
  out <-0
  for (i in seq(1,a,1)){
    out <- out + b^(i) 
  }
  return(out)
}

I would like to have test[,c=f(a,b)] gives :

test
# a b c
# 1 4 4
# 2 5 30    # 5 + 5^2
# 3 6 258   # 6 + 6^2 + 6^3

Is there a way to get the desired behaviour ?

Yes, this is where the problem stems from. However, for dt<-data.table(a=c(1,2)), calling the function g<-function(a){return(a)}this way dt[,b:=g(a)] leads to the desire result. The function g only take one parameter (the one of the row), and not the entire row. — kenshuri
– kenshuri, Commented Apr 21, 2020 at 13:36
Thanks again for your answer. The function I want to use is more complicated that this one, and it needs to use the elements of one column as number of loops to be done. I can't think of any workaround. For instance, with the input column 1,2,3 I would like the output column 1,2+2^2,3+3^2+3^3 calling the function f <- function(a){out<-0/n for (i in 1:a){out<- out + a^a}/n return(a)} — kenshuri
– kenshuri, Commented Apr 21, 2020 at 13:44
No, because this way, every row will get the same result. I will update the question with a more detailed explanation of the desired result. Thanks ! — kenshuri
– kenshuri, Commented Apr 21, 2020 at 13:57

kenshuri · Accepted Answer · 2020-04-22 08:56:54Z

0

Two solutions to solving the problem (thanks @chinsoon12) :

test[,c:=mapply(f, test[,a],test[,b])]

test[,c:=f(a,b),1L:nrow(test)]

Speed-wise, these two solutions are equivalent :

a<-1:500
b<-500:1

test_1 <- data.table(a,b)
test_2 <- data.table(a,b)

bench <- microbenchmark(v_1 = test_1[,c:=mapply(f,test_1[,a],test_1[,b])],v_2 = test_2[,c:=f(a,b),1L:nrow(test_2)],times=100L)

summary(bench)
#  expr      min       lq     mean   median       uq      max neval cld
#1  v_1 91.83598 95.63639 97.82780 96.94672 98.51073 113.2232   100   a
#2  v_2 91.72392 95.45878 98.92037 96.53573 98.71301 139.9906   100   a

autoplot(bench)

Benchmark plot

edited Apr 22, 2020 at 8:56

answered Apr 21, 2020 at 14:21

kenshuri

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1 Comment

chinsoon12 Over a year ago

another option is test[, v := f(a, b), 1L:nrow(test)] or in this case, test[, v := sum(cumprod(rep(b, a))), 1L:nrow(test)]

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