Order a list according to multiple columns in a dataframe in R

Question

I recently asked, and then answered, my own question after finding out it was a duplicate here:

For each named element in a list, return another named element from the same list based on closeness between element values

There, I used the eurodist dataset to find the closest neighbouring city Neigh of a city City based on the mean distance. I did this using split() along with lapply() .

library(data.table) # load package for transpose()

data(eurodist) # load eurodist data

labs <- labels(eurodist) # get city names
splt <- split(eurodist, labs) # split by city name

splt_mean <- lapply(splt, mean) # calculate mean for each city

x <- as.data.frame(splt_mean) # convert to data frame
x <- transpose(x) # transpose dataframe
colnames(x) <- "Mean" # name columns
rownames(x) <- labs # name rows

d <- data.frame(`diag<-`(as.matrix(dist(x$Mean)), Inf))
ids <- unlist(Map(which.min, d))
Neigh <- x$Mean[ids]
x <- data.frame(labs, x$Mean, Neigh) 
names(x)[1] <- "City"
names(x)[2] <- "Mean"
x[, 3] <- x$City[ids]

I've successfully applied the solution to my own data and now have one more step which I'm unable to figure out.

I'd like to order() splt so that corresponding row elements in City and Neigh occur together, City first followed by Neigh . For instance, calling the new list splt_sort , I need:

splt_sort
$Athens
[1] 3313 1326  966  330 1209 1418  328 2198 2250  618

$Rome
[1] 3927  204  747  789 1497  158  550 1178 2097 2707

...

Any thoughts?

Answer 1

I'll provide an answer to my own question, but @akrun deserves credit.

Their solution was a single line of R code:

splt2 <- splt[c(t(x[, c("City", "Neighbour")]))]

where x is subsetted to extract the c oncatenated vector comprising both City and Neighbour columns, then t ransposed and c oncatenated a final time before being applied to splt .