data.table update join by group - r

Question

I have a specific data.table question: is there a way to do an update join but by group ? Let me give an example:
df1 <- data.table(ID = rep(letters[1:3],each = 3),x = c(runif(3,0,1),runif(3,1,2),runif(3,2,3)))
df2 <- data.table(ID = c(letters[1],letters[1:5]))
> df2
ID
1: a
2: a
3: b
4: c
5: d
6: e
> df1
ID x
1: a 0.9719153
2: a 0.8897171
3: a 0.7067390
4: b 1.2122764
5: b 1.7441528
6: b 1.3389710
7: c 2.8898255
8: c 2.0388562
9: c 2.3025064
I would like to do something like
df2[df1,plouf := sample(i.x),on ="ID"]
But for each ID group, meaning that plouf would be a sample of the x values for each corresponding ID. The above line of code does not work this way, it sample the whole x vector:
> df2
ID plouf
1: a 1.3099715
2: a 0.8540039
3: b 2.0767138
4: c 0.6530148
5: d NA
6: e NA
You see that the values of plouf are not the x corresponding to the ID group of df1. I would like that the plouf value is between 0 and 1 for a, 1 and 2 for b, and 2 and 3 for c. I want to sample without replacement.
I tried :
df2[df1,plouf := as.numeric(sample(i.x,.N)),on ="ID",by = .EACHI]
which does not work:
Error in sample.int(length(x), size, replace, prob) :
cannot take a sample larger than the population when 'replace = FALSE'
This other attempt seems to be working:
df2$plouf <- df2[df1,on ="ID"][,sample(x,df2[ID == ID2,.N]),by = .(ID2 = ID)]$V1
But I find it hard to read or understand, it could be problematic for more than one grouping variable, and I am not sure it is quite efficient. I am sure there is a nice simple way to write it, but I don't have it. Any idea ?

Answer 1

Another option:
df1[df2[, .N, ID], on=.(ID), sample(x, N), by=.EACHI]
output:
ID V1
1: a 0.2655087
2: a 0.3721239
3: b 1.2016819
4: c 2.6607978
5: d NA
6: e NA
data:
library(data.table)
set.seed(0L)
df1 <- data.table(ID = rep(letters[1:3],each = 3),x = c(runif(3,0,1),runif(3,1,2),runif(3,2,3)))
df2 <- data.table(ID = c(letters[1],letters[1:5]))
Addressing comment:
library(data.table)
set.seed(0L)
df1 <- data.table(ID = rep(letters[1:3],each = 3),
NAME = rep(LETTERS[1:3],each = 3),
x = c(runif(3,0,1),runif(3,1,2),runif(3,2,3)))
df2 <- data.table(ID = c(letters[1],letters[1:5]),
NAME = c(LETTERS[1],LETTERS[1:5]))
df2[, ri := rowid(ID, NAME)][
df1[df2[, .N, .(ID, NAME)], on=.(ID, NAME), .(ri=1L:N, VAL=sample(x, N)), by=.EACHI],
on=.(ri, ID, NAME), VAL := VAL]
df2
If it is too repetitive to type ID, NAME, you can use
cols <- c("ID", "NAME")
df2[, ri := rowidv(.SD, cols)][
df1[df2[, .N, cols], on=cols, .(ri=1L:N, VAL=sample(x, N)), by=.EACHI],
on=c("ri", cols), VAL := VAL]
df2

Answer 2

Sample with replacement
You can do that like this:
df2[, plouf := df1[df2, on = .(ID),
sample(x, size = 1),
by=.EACHI]$V1]
You can join on the ID variable, but you must specify by=.EACHI as you are returning multiple values. The $V1 tells it to return the first column of the results.
Result:
ID sample
1: a 0.042188292
2: a 0.002502247
3: b 1.145714600
4: c 2.541768627
5: d NA
6: e NA
Sample without replacement
Its not pretty but it works:
df2$plouf = as.numeric(NA)
# create temporary table of number of sample required for each group
temp = df2[, .N, by = ID]
for(i in temp$ID){
# create a temporary sample
temp_sample = sample(df1[i==ID]$x, size = temp[ID==i]$n, replace = FALSE)
# assign sample
for(j in seq(1, length(temp_sample))){
df2[ID==i][j]$plouf = temp_sample[j]
}
}
Thanks to #David Arenburg for help