apply a function to a nested list

Question

I have already post a very similar question but the situation has changed and the answer that I got is not valid for these "new characteristics"

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 (e.g [[1]]$a=1, [[1]]$b=5, [[1]]$c=6, [[1]]d=1, then to the second ones (e.g [[1]]$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, e.g. 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

Although Reduce cannot be applied directly with Fisher.test it can be applied with cbind and then Fisher.test can be applied to that:

> lapply(lapply(pval.list, Reduce, f = cbind), apply, 1, Fisher.test)
[[1]]
[1] 1.953968e-03 2.999509e-05 5.320831e-04 1.339104e-02

[[2]]
[1] 0.0007878665 0.0052625525 0.0457435481 0.1146067577

[[3]]
[1] 8.982382e-03 3.055250e-08 1.064109e-02 5.094879e-02

[[4]]
[1] 9.911091e-03 2.032293e-06 8.073256e-03 1.458443e-02

[[5]]
[1] 2.357950e-03 6.135981e-11 3.326486e-01 6.038488e-01

[[6]]
[1] 6.597414e-03 3.470933e-09 3.362226e-01 5.708789e-01

ADDED:

This can be slightly shortened using simplify2array like this:

lapply(lapply(pval.list, simplify2array), apply, 1, Fisher.test)

or using data.frame like this:

lapply(lapply(pval.list, data.frame), apply, 1, Fisher.test)