In R, access indices stored in a list - r

Question

I have a data frame df, and a list L of indices at which I should put 0 instead of the current values of df.
Example:
DF:
# A tibble: 11 x 3
A B C
<dbl> <dbl> <dbl>
1724 4 2013
1758 4 2013
1612 3 2013
1692 3 2013
1260 33 2014
1157 22 2014
1359 63 2014
1414 27 2014
387 3 2016
374 3 2016
L:
[[1]]
[1] 3 4
[[2]]
[1] 1 2 3 4 5
[[3]]
[1] 1
So in this example, I have to put zeros in rows 3, 4 of column A, in rows 1:5 in column B and row 1 in column C.
Is there a way to do it as a one-liner in R? A dplyr or R-base solution would be great! Also, I would like to avoid apply or loops since I have to do this very efficiently

Answer 1

Loop looks very fast to me. Haven't done the complexity comparison but if you have your replacement in list form and want to replace with 'val', just simply:
df
a b c
1 1 1 1
2 2 2 2
3 3 3 3
4 4 4 4
5 5 5 5
6 6 6 6
7 7 7 7
8 8 8 8
9 9 9 9
10 10 10 10
val<-0
for(i in 1:length(L)){
df[L[[i]],i]<-val
}
df
a b c
1 1 0 0
2 2 0 2
3 0 0 3
4 0 0 4
5 5 0 5
6 6 6 6
7 7 7 7
8 8 8 8
9 9 9 9
10 10 10 10
I tested it on x, a 10,000 row and 10,0000 column df:
> b<-Sys.time()
> for(i in 1:length(L)){
+ x[L[[i]],i]<-0
+ }
> Sys.time()-b
Time difference of 0.490464 secs
Looks pretty quick :) I know it's obvious but hope it helps!
******** EDIT 1 ********
If we look at method by #mt1022 using unlist and cbind:
> b<-Sys.time()
> Lcol <- rep(seq_along(L), lengths(L))
> x[cbind(unlist(L), Lcol)] <- 0
> Sys.time()-b
Time difference of 7.467723 secs
Clearly much slower (because when we unlist, we essentailly loop through each and every element in L instead of each vector in L). ;)

Answer 2

Another way using matrix of indices:
# DF <- read.table(textConnection('A B C
# 1724 4 2013
# 1758 4 2013
# 1612 3 2013
# 1692 3 2013
# 1260 33 2014
# 1157 22 2014
# 1359 63 2014
# 1414 27 2014
# 387 3 2016
# 374 3 2016'), header = T)
#
# L <- list(c(3, 4), c(1, 2, 3, 4, 5), c(1))
Lcol <- rep(seq_along(L), lengths(L))
DF[cbind(unlist(L), Lcol)] <- 0
# > DF
# A B C
# 1 1724 0 0
# 2 1758 0 2013
# 3 0 0 2013
# 4 0 0 2013
# 5 1260 0 2014
# 6 1157 22 2014
# 7 1359 63 2014
# 8 1414 27 2014
# 9 387 3 2016
# 10 374 3 2016

Answer 3

Another option is to use mapply in combination with do.call.
do.call(cbind, mapply(function(x,y){
df[x,y]<-0
df[y]
}, mylist, seq_along(mylist)))
# A B C
# [1,] 1724 0 0
# [2,] 1758 0 2013
# [3,] 0 0 2013
# [4,] 0 0 2013
# [5,] 1260 0 2014
# [6,] 1157 22 2014
# [7,] 1359 63 2014
# [8,] 1414 27 2014
# [9,] 387 3 2016
# [10,] 374 3 2016
Data:
df <- read.table(text =
"A B C
1724 4 2013
1758 4 2013
1612 3 2013
1692 3 2013
1260 33 2014
1157 22 2014
1359 63 2014
1414 27 2014
387 3 2016
374 3 2016", header = TRUE)
mylist <- list(c(3, 4), c(1, 2, 3, 4, 5), c(1))