在嵌套列表的不同级别应用函数（重新发布）

Question

Older post:

I am wondering if it is possible to apply a function in different levels of a nested list, rapply applies a function recursively in a list but in the same level. My question is related to apply a function in different levels with different lengths. An example for illustration:

list <- list(list(a=1:5, b=5:9, c=6:10, d=1:5),
             list(e=2:6, f=3:7, g=8:12),
             list(h=3:7, i=6:10, j=11:15, k=2:6),
             list(l=4:8, m=2:6),
             list(n=5:9, o=1:5, p=2:6, q=0:4),
             list(r=6:10, s=3:7, t=9:13))

I would like to apply a function, such as sum , to the first elements (eg [[1]]$a=1, [[1]]$b=5, [[1]]$c=6, [[1]]d=1 , then to the second ones (eg [[1]]$a=2, [[1]]$b=6, [[1]]$c=7, [[1]]$d=2) and so on. The result should be something like that:

[[1]]
13 17 21 25 29

[[2]]
13 16 19 22 25

[[3]]
22 26 30 34 38

[[4]]
6 8 10 12 14

[[5]]
8 12 16 20 24

[[6]]
18 21 24 27 30

Perhaps a combination of rapply and mapply ?

Thanks

New post :

@G.Grothendieck already gave me a good solution for this chase however, I have other list with pvalues where I would like to apply more complex functions, eg mean or other functions such as:

Fisher.test <- function(p) {
  Xsq <- -2*sum(log(p))
  p.val <- 1-pchisq(Xsq, df = 2*length(p))
  return(p.val)
}

Reduce does not work as it does with functions like sum or with f="+" , any suggestions??

Here is an example how this new list looks like"

pval.list <- list(list(a=c(0.05, 0.0001, 0.32, 0.45), b=c(0.1,0.12,0.01,0.06), c=c(0.1,0.12,0.01,0.06), d=c(0.01,0.02,0.03,0.04)),
             list(e=c(0.04, 0.1, 0.232, 0.245), f=c(0.05, 0.01, 0.22, 0.54), g=c(0.005, 0.1, 0.032, 0.045)),
             list(h=c(0.03, 0.01, 0.12, 0.4), i=c(0.5, 0.0001, 0.132, 0.045), j=c(0.005, 0.0001, 0.0032, 0.045), k=c(0.5, 0.1, 0.932, 0.545)),
             list(l=c(0.022, 0.0012, 0.32, 0.45), m=c(0.0589, 0.0001, 0.0032, 0.0045)),
             list(n=c(0.051, 0.01, 0.32, 0.45), o=c(0.05, 0.0001, 0.32, 0.45), p=c(0.05, 0.0001, 0.32, 0.45), q=c(0.05, 0.0001, 0.32, 0.45)),
             list(r=c(0.053, 0.001, 0.32, 0.45), s=c(0.05, 0.0001, 0.32, 0.45), t=c(0.05, 0.0001, 0.32, 0.45)))

Answer 1

Try this:

> lapply(list, Reduce, f = "+")
[[1]]
[1] 13 17 21 25 29

[[2]]
[1] 13 16 19 22 25

[[3]]
[1] 22 26 30 34 38

[[4]]
[1]  6  8 10 12 14

[[5]]
[1]  8 12 16 20 24

[[6]]
[1] 18 21 24 27 30