select one element of a nested list and convert it to dataframe in R

Question

I have the following list structure

ts1 <- ts(rnorm(10))
l <- list()
l[["P1"]][["M1"]][["forecast"]] <- rnorm(1)
l[["P1"]][["M1"]][["accuracy"]] <- rnorm(1)
l[["P1"]][["M1"]][["forecast"]] <- rnorm(1)
l[["P1"]][["M1"]][["accuracy"]] <- rnorm(1)
l[["P2"]][["M2"]][["forecast"]] <- rnorm(1)
l[["P2"]][["M2"]][["accuracy"]] <- rnorm(1)
l[["P2"]][["M2"]][["forecast"]] <- rnorm(1)
l[["P2"]][["M2"]][["accuracy"]] <- rnorm(1)

I wish to convert it to a data frame with the following columns: Note that I only want one of the elements of the final list

df <- data.frame(Product=c("p1","p1","p2","p2"),model=c("M1","M2","M1","M2"),accuracy=c(l$P1$M1$accuracy,l$P1$M2$accuracy,l$P2$M1$accuracy,l$P2$M2$accuracy))

I tried something like this:

df <- lapply(l, function(x) do.call(rbind,x))
df <- do.call(rbind,Map(cbind,product=names(df),df))
as.data.frame(df)

But I don´t manage to get the names properly

Ronak Shah · Accepted Answer · 2020-10-11 13:16:23Z

1

In base R, you could try something like this :

tmp <- lapply(l, function(x) {
  df <- do.call(rbind,x)
  data.frame(df, model = rownames(df))
})
do.call(rbind, Map(cbind, tmp, Product = names(tmp)))

#     forecast   accuracy model Product
#P1 -0.6148388 -0.6326906    M1      P1
#P2  0.1245955  0.9269495    M2      P2

However, it is shorter with tidyverse functions.

library(dplyr)
library(purrr)

map_df(l, ~bind_rows(.x, .id = 'model'), .id = 'product')

#  product model forecast accuracy
#  <chr>   <chr>    <dbl>    <dbl>
#1 P1      M1      -0.615   -0.633
#2 P2      M2       0.125    0.927

answered Oct 11, 2020 at 13:16

Ronak Shah

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akrun · Accepted Answer · 2020-10-11 20:51:45Z

1

An option with data.table

library(data.table)
rbindlist(lapply(l, rbindlist, idcol = 'model'), idcol = 'product')
#   product model   forecast  accuracy
#1:      P1    M1 0.72132201 0.9557613
#2:      P2    M2 0.01656428 0.8460819

answered Oct 11, 2020 at 20:51

akrun

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