Applying a function across nested list

问题

Say, I have the following list

raw <- list(list(1:2, 2:3, 3:4), list(4:5, 5:6, 6:7), list(7:8, 8:9, 9:10))

I would like to find the mean of the corresponding entries of the out-most list. The expected output would be something like

[[1]]
[1] 4 5

[[2]]
[1] 5 6

[[3]]
[1] 6 7

This is because the mean of 1:2, 4:5, and 7:8 would be 4:5.

I have been experimenting with stuff like lapply(raw, function(x) lapply(x, mean)), but apparently it doesn't return the desired output.

回答1:

This is pretty ugly, but we can use mapply to iterate over the lists but we need to expand the list into parameters via do.call

do.call("mapply", c(function(...) rowMeans(data.frame(...)), raw, SIMPLIFY=FALSE))

You can make this prettier using the purrr package

purrr::pmap(raw, ~rowMeans(data.frame(...)))

回答2:

n = length(raw[[1]])
lapply(1:n, function(i){
    d = do.call(rbind, lapply(seq_along(raw), function(j){
        raw[[j]][[i]]
    }))
    apply(d, 2, mean)
})
#[[1]]
#[1] 4 5

#[[2]]
#[1] 5 6

#[[3]]
#[1] 6 7

aggregate(. ~ ind, do.call(rbind, lapply(raw, function(x)
    data.frame(cbind(do.call(rbind, x), ind = seq_along(x))))), mean)
#  ind V1 V2
#1   1  4  5
#2   2  5  6
#3   3  6  7

回答3:

You could put the thing into an array and take the cell medians (I suppose you want these instead of means).

A <- array(matrix(unlist(raw), 2, byrow=FALSE), dim=c(2, 3, 3))
v.mds <- t(apply(A, 1:2, median))
lapply(1:3, function(x) v.mds[x, ])
#   [[1]]
# [1] 4 5
# 
#   [[2]]
# [1] 5 6
# 
#   [[3]]
# [1] 6 7

Generalized like so:

A <- array(matrix(unlist(raw), length(el(el(raw))), byrow=0), 
           dim=c(length(el(el(raw))), el(lengths(raw)), length(raw)))
v.mds <- t(apply(A, 1:2, median))
lapply(1:nrow(v.mds), function(x) v.means[x, ])

来源：https://stackoverflow.com/questions/58453574/applying-a-function-across-nested-list

标签

list

nested

apply

lapply