Access lapply index names inside FUN

Question

Is there a way to get the list index name in my lapply() function?

n = names(mylist)
lapply(mylist, function(list.elem) { cat(\"What is the name of this list         


        

Answer 1

@ferdinand-kraft gave us a great trick and then tells us we shouldn't use it because it's undocumented and because of the performance overhead.

I can't argue much with the first point but I'd like to note that the overhead should rarely be a concern.

let's define active functions so we don't have to call the complex expression parent.frame()$i[] but only .i(), We will also create .n() to access the name, which should work for both base and purrr functionals (and probably most others as well).

.i <- function() parent.frame(2)$i[]
# looks for X OR .x to handle base and purrr functionals
.n <- function() {
  env <- parent.frame(2)
  names(c(env$X,env$.x))[env$i[]]
}

sapply(cars, function(x) paste(.n(), .i()))
#>     speed      dist 
#> "speed 1"  "dist 2"

Now let's benchmark a simple function Which pastes the items of a vector to their index, using different approaches (this operations can of course be vectorized using paste(vec, seq_along(vec)) but that's not the point here).

We define a benchmarking function and a plotting function and plot the results below :

library(purrr)
library(ggplot2)
benchmark_fun <- function(n){
  vec <- sample(letters,n, replace = TRUE)
  mb <- microbenchmark::microbenchmark(unit="ms",
                                      lapply(vec, function(x)  paste(x, .i())),
                                      map(vec, function(x) paste(x, .i())),
                                      lapply(seq_along(vec), function(x)  paste(vec[[x]], x)),
                                      mapply(function(x,y) paste(x, y), vec, seq_along(vec), SIMPLIFY = FALSE),
                                      imap(vec, function(x,y)  paste(x, y)))
  cbind(summary(mb)[c("expr","mean")], n = n)
}

benchmark_plot <- function(data, title){
  ggplot(data, aes(n, mean, col = expr)) + 
    geom_line() +
    ylab("mean time in ms") +
    ggtitle(title) +
    theme(legend.position = "bottom",legend.direction = "vertical")
}

plot_data <- map_dfr(2^(0:15), benchmark_fun)
benchmark_plot(plot_data[plot_data$n <= 100,], "simplest call for low n")

benchmark_plot(plot_data,"simplest call for higher n")

^{Created on 2019-11-15 by the reprex package (v0.3.0)}

The drop at the start of the first chart is a fluke, please ignore it.

We see that the chosen answer is indeed faster, and for a decent amount of iterations our .i() solutions are indeed slower, the overhead compared to the chosen answer is about 3 times the overhead of using purrr::imap(), and amount to about, 25 ms for 30k iterations, so I lose about 1 ms per 1000 iterations, 1 sec per million. That's a small cost for convenience in my opinion.