Sum specified elements of sub-lists - r

Question

The idea:
Suppose I have a lists with two vectors. Then, I would like to take the first element of the first vector and divide it by the sum of it and the first element of the second vector of the list. Then do that for all elements of the first list. After that, do the same thing but with the second vector of the list.
The code of the lists:
tau1 <- list(c(0.43742669 , 0.64024429, 0.39660069, 0.11849773), c(0.5060767, 0.4857891, 0.4553237, 0.5045598))
My worked code for only two vectors.
Tau1 <- vector('list', 2)
for(i in seq_along(tau1)){
for(j in 1:length(tau1[[1]])){
Tau1[[i]][[j]] <- tau1[[i]][[j]] / Reduce('+', tau1[[1]][[j]], tau1[[2]][[j]])
}
}
Example:
First element of the list:
TT1 <- tau1[[1]][[1]]/(tau1[[1]][[1]]+tau1[[2]][[1]])
[1] 0.4636196
Then for the second element of the list:
TT2 <- tau1[[2]][[1]]/(tau1[[1]][[1]]+tau1[[2]][[1]])
[1] 0.5363804
The problem:
I would like to do that for arbitrary number of vectors. For example,
Reduce('+', tau1[[1]][[j]], tau1[[2]][[j]], tau1[[3]][[j]], tau1[[4]][[j]])
How can I do that automatically? any help, please?

Answer 1

If we are using Reduce, then we need to remove the [[i]] to get the sum of corresponding list elements to get a vector. Then subset by the 'j'th index to divide the 'j'th element of 'tau1[[i]]'
Tau1 <- vector('list', 2)
for(i in seq_along(tau1)){
for(j in seq_along(tau1[[1]])){
Tau1[[i]][[j]] <- tau1[[i]][[j]] /Reduce(`+`, tau1)[j]
}
}
Regarding the error mentioned in the comments, it can happen if there are non-numeric elements. The OP mentioned about NULL elements, but NULL occurs as a single element in a list. So, there is a possibility of character "NULL". For e.g.
tau1 <- list(c(0.43742669 , 0.64024429, "NULL", 0.11849773),
c(0.5060767, 0.4857891, 0.4553237, 0.5045598))
Upon running the code above
Error in f(init, x[[i]]) : non-numeric argument to binary operator

Answer 2

To work with any number of list elements use apply family of functions with [[.
To extract j element in each sub-list use sapply(tau1, "[[", j). To sum those elements use: sum(sapply(tau1, "[[", j))
PS.: Instead of for(j in 1:length(tau1[[1]]){} you should have for(j in 1:length(tau1[[i]]){} - just in case.

Answer 3

Here is a base R one-liner:
lapply(tau1, "/", do.call(mapply, c(FUN = sum, tau1)))
# [[1]]
# [1] 0.4636196 0.5685838 0.4655351 0.1901875
#
# [[2]]
# [1] 0.5363804 0.4314162 0.5344649 0.8098125
Or alternatively (from #lmo's comment):
lapply(tau1, "/", Reduce("+", tau1))
Here is a purrr equivalent:
library(purrr)
tau1 %>% map(`/`, pmap_dbl(., sum))
# [[1]]
# [1] 0.4636196 0.5685838 0.4655351 0.1901875
#
# [[2]]
# [1] 0.5363804 0.4314162 0.5344649 0.8098125