'Calculating mean of each data frame inside a list of lists and assigning it to a new vector in R

This is my data:

df1 <- data.frame(x = 1:5, y = letters[1:5])
df2 <- data.frame(x = 1:15, y = letters[1:15])
df3 <- data.frame(x = 1:25, y = letters[1:25])
df4 <- data.frame(x = 1:6, y = letters[1:6])
df5 <- data.frame(x = 1:8, y = letters[1:8])

l1 <- list(df1, df2)
l2 <- list(df3, df4, df5)
mylist <- list(l1, l2)

I want to calculate the mean of the x column in all data frames inside mylist, and put them in a new empty list (or vector), like so:

mean_vec <- c(
 mean(df1$x),
 mean(df2$x),
 mean(df3$x),
 mean(df4$x),
 mean(df5$x)
)
rlistfunctionlapplymean


Solution 1:[1]

Another possible solution, based on purrr::map_depth:

library(tidyverse)

map_depth(mylist, 2, ~ mean(.x$x)) %>% unlist

#> [1]  3.0  8.0 13.0  3.5  4.5

Or using rrapply::rrapply, solution that is now shorter thanks to @Maël's comment, to whom I thank:

library(rrapply)
library(magrittr)

rrapply(mylist, condition = is.numeric, f = mean, how = "unlist") %>% unname

#> [1]  3.0  8.0 13.0  3.5  4.5

Solution 2:[2]

You can unlist your nested list and compute the mean for each:

mean_vec <- sapply(unlist(mylist, recursive = F), function(dat) mean(dat$x))

mean_vec
# [1]  3.0  8.0 13.0  3.5  4.5

Another option with rapply:

mean <- rapply(mylist, mean)
unname(mean[names(mean) == "x"])
# [1]  3.0  8.0 13.0  3.5  4.5

Solution 3:[3]

A purrr solution

library(purrr)
library(dplyr)

mylist %>%
  map_depth(., 2, ~ .x %>% summarise(mean = mean(x, na.rm = T))) %>%
  bind_rows() %>%
  pull()

Sources

This article follows the attribution requirements of Stack Overflow and is licensed under CC BY-SA 3.0.

Source: Stack Overflow

Solution Source
Solution 1
Solution 2
Solution 3 Julian

Related Questions